Consider the set $S \subset [0, \ 1]$ defined as: $$S = \Big\{ 1 - \frac{1}{n} \ \biggr\rvert \ n \in \mathbb{Z}^{+} \Big\} = \Big\{ 0, \ \frac{1}{2}, \ \frac{2}{3}, \ ... \Big\}$$ Let us pick some real $s \in [0, \ 1)$. First, we note that since $s < 1$, $1 - s$ is also a real number, and so is $1/(1 - s)$. Using the Archimedean property, we can then pick an integer $n$ such that $1/(1 - s) \leq n$. Using the well-ordering principle, the set of all $n$ that satisfy this inequality has a minimal element, which we call $m$. Clearly, it then holds true that: $$m - 1 \leq \frac{1}{1 - s} \leq m$$ so we will have: $$1 - \frac{1}{m - 1} \leq s \leq 1 - \frac{1}{m}$$ This implies that $s$ lies between two elements of the set $S$. Since $[0, \ 1]$ is a linear continuum, there is an element of $[0, \ 1]$ between $s$ and $1 - 1/m$, which we call $y$. We then consider the open set $[s, \ y)$. We have found, for each point of $[0, \ 1)$, an open set containg this point that does not intersect $S$ at a point other than itself. It follows that $S$ has no limit point in $[0, \ 1)$. In addition, we can consider the open set $\{1\}$ containing $1$, which clearly does not intersect $S$.

Thus, we have found an infinite set $S$, such that every point of $[0, \ 1]$ is contained in an open set that does not intersect $S$ at a point other than itself. It follows that $S$ has no limit points, and is thus not limit point compact.

Let us first assume that the space is sequentially compact. We will assume that there exists a countable covering, $\{U_n\}$ that does not have a finite subcover.

We define a sequence by setting: $$x_i \notin U_1 \cup U_2 \cup ... \cup U_i$$ We can do this as we have assumed there is no finite subcover of $\{U_n\}$. We know that there must exist some $x$ to which a subsequence of $x_i$ converges to. By definition, $x \in U_k$ for some $k$. Except, for all $i \geq k$, we know that $x_i \notin U_k$, so the points can't converge to $x$. Thus, we have a derived a contradiction, and the space must be countably compact.

First assume that the space is countably compact. Let us consider some nested sequence, $C_1 \supset C_2 \supset \cdots$. It must be true that: $$\displaystyle\bigcap_{n \in \mathbb{Z}^+} = \displaystyle\bigcap_{n \in \mathbb{Z}^+} (X - U_n) = X - \displaystyle\bigcup_{n \in \mathbb{Z}^+} U_n$$ where each $U_n$ is open in $X$. Assume that $\cup_{n} U_n = X$. It would then follow that $\{U_n\}$ is a countable cover of $X$, so it contains a finite subcover that also covers $X$. Let $k$ be the largest index of the subcover. Clearly, for $i < j$, we have $U_i \subset U_j$, so it follows that the union of the elements of the subcover is equal to $U_k$. It then must be true that $U_k = X$, which would imply that $C_k = X - U_k = \emptyset$, a contradiction, as we have assumed each $C_n$ to be nonempty.

Thus, it follows that $\cup_{n} U_n \neq X$, and $X - \cup_{n} U_n \neq \emptyset$. Therefore, the intersection of the nested sequence is non-empty.

Now, let us assume that each nested sequence of closed sets has a non-empty intersection. Consider the countable cover $\{U_n\}$ of $X$. Assume that each closed set: $$C_n = X - \displaystyle\bigcup_{k = 1}^{n} U_k$$ is non-empty. It follows that: $$\displaystyle\bigcap_n C_n = \displaystyle\bigcap_n \Big[ X - \displaystyle\bigcup_{k = 1}^{n} U_k \Big] = X - \displaystyle\bigcup_{n} \displaystyle\bigcup_{k = 1}^{n} U_k$$ is non-empty. Except we know that $\{U_n\}$ covers $X$, so $\cup_{n} \cup_{k = 1}^{n} U_k = \cup_n U_n = X$. Thus the intersection is empty, contradicting the initial assumption that any such intersection is non-empty. It follows that there must exist some $m$ such that: $$C_n = X - \displaystyle\bigcup_{k = 1}^{m} U_k$$ is empty, implying that $\cup_{k = 1}^{m} U_k = X$, so we have a finite subcover that covers the space.

First, let us show that $f$ is a bijection. Pick $x$ and $y$ of $X$ such that $f(x) = f(y)$. It follows that: $$d(x, \ y) = d(f(x), \ f(y)) = d(f(x), \ f(x)) = 0$$ so by definition of the metric, $x = y$. Thus, the map is injective. Now, we prove surjectivity. Assume that there is some $a \in X$ such that $a \notin f(X)$. Recall that $X$ is compact, so $f(X)$ is compact. It follows that $d(a, \ f(X)) = d(a, \ f(x))$ for some $f(x) \in f(X)$ (we proved this result in the exercises of the last section). Since $a \neq f(x)$, it follows that: $$d(a, \ f(x)) = \epsilon \ \Rightarrow \ B_d(a, \ \epsilon) \cap f(X) = \emptyset$$ so there is an $\epsilon$-neighburhood around $a$ that is disjoint from $f(X)$.

Now, we define a sequence by setting $x_1 = a$ and $x_{n + 1} = f(x_n)$. Let us pick $n$ and $m$ such that $n \neq m$ (without loss of generality, assume $n < m$). We will have: $$d(x_n, \ x_m) = d(x_n, \ f \cdots f(x_n))$$ where the function $f$ is applied $m - n$ times. Using the original formula, we then have: $$d(x_n, \ f \cdots f(x_n)) = d(a, \ f \cdots f(a))$$ where the function is repeated $m - n$ times. Clearly, the second argument in the metric map is an element of $f(X)$. We have assumed that $a$ is contained in an $\epsilon$-neighbourhood disjoint from $f(X)$, so: $$\epsilon \leq d(a, \ f \cdots f(a)) = d(x_n, \ x_m)$$ for all $n$ and $m$ with $n \neq m$.

Now, we complete the proof. Since $X$ is compact and a metric space, it will also be sequentially compact. Assume that there exists a finite subsequence of $x_i$ given by $x_{n_i}$ that converges to some point $x$. It follows that for the $\epsilon/2$-ball around $x$, there exists some $N$ such that for $n_i \geq N$, we have $x_{n_i} \in B_{\epsilon/2}$. But this would imply that there exist $x_n$ and $x_m$ such that $d(x_n, \ x_m) < \epsilon$, a clear contradiction.

It follows that no such $a$ can exist, and $f(X) = X$. Thus, the function is surjective.

The remainder of the proof is straightforward: we simply note that given some function $f$ that is an isometry, $f(B(x, \ \epsilon)) = B(f(x), \ \epsilon)$, and we apply this fact to both $f$ and $f^{-1}$ showing continuity both ways.

Let us first assume that there exist two fixed points, $x$ and $y$. It follows that $d((f(x), \ f(y)) = d(x, \ y)$. But $f$ is a contraction, so there must exist $\alpha < 1$ such that: $$d(f(x), \ f(y)) = d(x, \ y) \leq \alpha d(x, \ y)$$ Clearly, no such $\alpha$ exists. It follows that there is at most one fixed point of $f$.

Now, let us prove that $f$ is continuous. We pick some $f(x) \in X$ and an open set $V$ around $f(x)$. Since $X$ is a metric space, there is an $\epsilon$ ball around $f(x)$ that is contained in $V$. Since $f$ is a contraction, we know that a number $\alpha$ exists that satisfies the criteria in Definition 3.

Consider the open ball $B(x, \ \epsilon)$. Let $y$ be an element of the open ball. It is then true that $d(x, \ y) < \epsilon$. We then know that $d(f(x), \ f(y)) \leq \alpha \epsilon < \epsilon$, for $f(y)$ is in the $\epsilon$-ball around $f(x)$. It follows that $f(B) \subset V$. This proves continuity.

Now, consider the sets of the form $A_n = f^{n}(X)$ where $f^1 = f$ and $f^{n + 1} = f \circ f^n$. Since $f$ is continuous, each $A_n$ will be compact. In addition, we are operating in a metric space, so it is automatically Hausdorff. This implies that each $A_n$ is closed. Clealy, $A_{n + 1} \subset A_n$, so it follows that the set of all $A_n$ has the finite intersection property. This implies that: $$A = \displaystyle\bigcap_{n \in \mathbb{Z}^+} A_n$$ is non-empty. It is clear that the set of fixed points will be contained in $A$. In addition, if we assume that there is a point $a$ of $A$, such that $a$ is not a fixed point,

First, let us assume that there are two fixed points $x, y \in X$. It would then be true that: $$d(f(x), \ f(y)) < d(x, \ y) \ \Rightarrow \ d(x, \ y) < d(x, \ y)$$ a clear contradiction. Thus, there is at most one fixed point of the space. Now, we pick some open $V$ of $X$ around some $f(x)$. There is an $\epsilon$-ball around $f(x)$. Now, consider the $\epsilon$ ball around $x$, which we call $B_x$. For some $y \in B_x$, we will have $d(f(x), \ f(y)) < d(x, \ y) < \epsilon$, so $f(y)$ is in the $\epsilon$ ball around $f(x)$. It follows that $f$ is continuous.

With this fact, we then use the same reasoning from above to conclude that $A$ is non-empty (we define it in the same way). We pick some $x \in A$ (we know that $A$ is non-empty, from above). We now define the sequence $x_n$ so that: $$f^{n + 1}(x_n) = x$$ we then define another sequence: $y_n = f^{n} (x_n)$. Since $X$ is a compact metric space, it is sequentially compact. It follows that $y_n$ has a convergent subsequence that converges to some point $a$. We index this subsequence as $y_{n_i}$. If we assume that there exist some $n$ such that $a \notin f^n(A)$, we can employ the same strategy in the above exercises to define an $\epsilon$-neighbourhood around $a$. Clearly, for such an open set, no $N$ exists such that for $n_k \geq N$, $y_{n_k}$ is contained in the $\epsilon$ neighbourhood, as any $y_{n_k}$ for $n_k \geq n$ will lie inside $f^n(A)$, and thus outside the open neighbourhood. It follows that $a \in A$.

Now, since $y_{n_i} \ \rightarrow \ a$, then $f(y_{n_i}) \ \rightarrow \ f(a)$, so we get: $$f(y_{n_i}) = (f \circ f^{n_i})(x_{n_i}) = f^{n_i + 1}(x_{n_i}) = x$$ so it follows that the sequence given by taking the continuous image of each $y_{n_i}$ is the constant sequence, $x$, so $f(a) = x$.

This result implies that some $x \in A$ can be written as $f(a) \in f(A)$, so $A \subset f(A)$. We also note that $f(A) \subset A$. Thus, $f(A) = A$.

We then have: $$\text{diam} \ A = \sup\{d(x, \ y) \ | \ x, y \in A\} = \sup\{d(f(x), \ f(y)) \ | \ x, y \in A\} \leq \alpha \sup\{d(x, \ y) \ | \ x, y \in A\}$$ so since $\alpha < 1$, it follows that $\text{diam} \ A = 0$. This implies that the set $A$ contains one point.

Now, we know that $A$ contains only one point. We have shown that $x \in A$ and $a \in A$, so it must be true that $x = a$. We also know that $f(a) = x$, so it follows that $f(x) = x$ and $x$ is a fixed point.

We have shown that the space contains at most one fixed point, and at least one fixed point. Thus, the space contains exactly one, unique fixed point, and the proof is complete.

Let us pick two distinct points $x$ and $y$ of $[0, \ 1]$. Without loss of generality, assume that $x < y$. We then have: $$d(f(x), \ f(y)) = \Big| y - y^2/2 - x + x^2/2 \Big|$$ we know that $y - x$ is positive. In addition, we see that since $x < y$, $x^2 < y^2$, so: $$d(f(x), \ f(y)) = \Big| y - y^2/2 - x + x^2/2 \Big| < |y - x| = d(x, \ y)$$ Thus, the map is a shrinking map. Now, assume that there exists some $\alpha \in [0, \ 1)$ such that: $$|f(x) - f(y)| \leq \alpha d(x, \ y)$$ for all $x$ and $y$ in $[0, \ 1]$. It follows that: $$\frac{|f(x) - f(y)|}{|x - y|} \leq \alpha$$ From the mean value theorem of calculus, we know that given $x$ and $y$, there exists some $z \in (x, \ y)$ such that: $$f'(z) = 1 - z = \frac{|f(x) - f(y)|}{|x - y|} \leq \alpha$$ Let us pick $x = 0$ and $y = 1 - \alpha$. We then must have $z \in (0, \ 1 - \alpha)$, so $z < 1 - \alpha$, which implies that $1 - z > \alpha$. But this is clearly a contradiction to the above equation. Thus, no such $\alpha$ exists, and the map is not a contraction.

Now, we show that $f$ is not a contraction. Assume that some $\alpha$ exists such