Consider some subset $S$ of $X$ that has an upper bound in $X$. Let $L$ be the set of all such upper bounds (it will be non-empty, by the initial assumption). For each $s \ \in \ S$, consider the closed set: $$C_s \ = \ \{x \ | \ s \ \leq \ x\}$$ Let $C$ be the intersection of all such sets. This will be the set $L$, as both are the sets of all $x$ such that $s \ \leq \ x$ for all $s \ \in \ S$. It follows that $L$ is closed.

Assume that $L$ has no least element. It follows that the open sets $\{(\ell, \ \infty) \ | \ \ell \ \in \ L\}$ cover $L$. Consider a finite subcollection of such sets, $\{(\ell_1, \ \infty), \ ..., \ (\ell_n, \ \infty)\}$. Let $\ell_i$ be the minimum of all such $\ell$ values. By definition, $\ell_i \ \in \ L$ but is not in the union of the open sets. This contradicts the fact that $L$ is compact, which implies that $L$ must have a minimal element. This is will be the least upper bound of $S$.

Consider some non-empty subset $S$ of $X$ that has an upper bound. We let $L$ be the set of upper bounds of $S$. Let us consider $s \in S$ and $\ell \in S$. If $s = \ell$, then this is automatically the least upper bound of $S$ and we are done. Otherwise, consider the closed interval $[s, \ \ell]$.

Assume that $x \ \in \ \bar{A}$. It follows that every open set of $x$ intersects $A$. Thus, for any $\epsilon \ > \ 0$, we can choose some open $B_{\epsilon}(x)$ such that it contains an element of $a$, implying that $d(x, \ a) \ < \ \epsilon$. Assume that $f \ = \ \inf \{d(x, \ a) \ | \ a \ \in \ A\}$ is greater than $0$. It follows that there exists some $\delta$ such that $0 \ < \ \delta \ < \ f$ such that $d(x, \ a) \ > \ \delta$ for all $a \ \in \ A$. This contradicts the prior statement, which implies that $f \ = \ 0$.

Conversely, assume that $d(x, \ A) \ = \ 0$. Also assume that $x \ \notin \ \bar{A}$. This means that there exists some open set around $U$ that does not intersect $A$. By definition of the metric topology, there also exists some $\epsilon$-ball $B$ around $x$ such that $B \ \subset \ U$. Thus, $d(x, \ a) \ \geq \ \epsilon$ for any $a \ \in \ A$, which is a contradiction. This implies that $x$ must be in $\mathcal{A}$.

This is just another way of saying the set $\{d(x, \ a) \ | \ a \ \in \ A\}$ has a smallest element. We know that the function $d : X \ \times \ X \ \rightarrow \ \mathbb{R}$ is continuous. If we restrict the domain to $x \ \times \ A$, we also have a continuous function. Clearly, this new domain is compact, as $A$ is compact. Thus, by extreme value theorem, there exists a minimum value of $d(x, \ a')$ Thus, $d(x, \ A) \ = \ \inf \ \{d(x, \ a) \ | \ a \ \in \ A\} \ = \ d(x, \ a')$, and the theorem is proved.

Pick some $x \ \in \ U(A, \ \epsilon)$. It follows that $d(x, \ A) \ < \ \epsilon$. Assume that there exists no $d(x, \ a') \ < \ \epsilon$. This contradicts the assumption that $d(x, \ A)$ is the greatest lower bound of the set $\{d(x, \ a) \ | \ a \ \in \ A\}$, as $\epsilon$ would be a greater lower bound. Thus, there exists some $d(x, \ a') \ < \ \epsilon$, which implies that $x \ \in \ B_d(a', \ \epsilon)$, which is in the union of all such sets.

Now, assume that $x \ \in \ B_{\epsilon}$ (the union of all $\epsilon$-balls around each $a \ \in \ A$). Consider the set of all $\epsilon$-balls centred at $a$ that contain $x$. It must be true that $d(x, \ a) \ < \ \epsilon$ for all such balls. As for the balls that don't contain $x$, it follows that $d(x, \ a) \ \geq \ \epsilon$. Thus, it follows that $\inf \ \{d(x, \ a) \ | \ a \ \in \ A\} \ < \ \epsilon$, which implies that $x \ \in \ U(A, \ \epsilon)$. It follows that $B_{\epsilon} \ = \ U(A, \ \epsilon)$.

This implies that we should be able to choose some $\epsilon$ such that all $\epsilon$-balls around $a \ \in \ A$ are in $U$. Since $U$ is open, for any $x \ \in \ U$ (and thus any $a \ \in \ A$), we can define an $\epsilon_a$-ball around $a$ that that $B_{\epsilon_a}(a) \ \subset \ U$.

Consider the metric $d$ as a map $d : X \ \times \ X \ \rightarrow \ \mathbb{R}$, restricted to $d(X \ \times \ X)$. Assume that $d(X \ \times \ X) \ \neq \ [a, \ b]$, for some $a$ and $b$. It follows that for some two points of $d(X \ \times \ X)$, which we call $x$ and $y$, with $x \ < \ y$, we can choose some $z$ with $x \ < \ z \ < \ y$ such that $z \ \notin d(X \ \times \ X)$. We know that $X$ has more than one point, so there are at least two points of $d(X \ \times \ X)$ which can take the place of $x$ and $y$ (namely, $d(p, \ p)$ and $d(p, \ q)$, for $p, \ q \ \in \ X$).

Now, let us consider the open sets $U \ = \ (-\infty, \ z) \ \cap \ d(X \ \times \ X)$ and $V \ = \ (\infty, \ z) \ \cap \ d(X \ \times \ X)$ of $d(X \ \times \ X)$. Clearly: $$U \ \cup \ V \ = \ d(X \ \times \ X)$$ Since $d$ is continuous and bijective, it follows that $d^{-1}(U)$ and $d^{-1}(V)$ are open, disjoint sets whose union is $X$. This contradicts the fact that $X$ is connected. Thus, $d(X \ \times \ X)$ is an interval in the real numbers. We have proved that any interval of the reals is uncountable, and since $d$ is a bijection, $X \ \times \ X$ is also uncountable.

Assume that $X$ is countable. It follows that $X \ \times \ X$ is countable, a contradiction. Thus, $X$ is uncountable.

Let us consider some subset $S \subset C$. If $S$ has one point, it will be connected, which is allowed. Assume that $S$ has at least two points, $a$ and $b$. We wish to show that there exists a "deleted section" that lies between $a$ and $b$, thus allowing us to form a separation. First of all, consider the "distance" from $a$ to $b$, which is given by $|b - a|$. We assume without loss of generality that $b > a$.

Since $b - a$ is a real number, by the Archimedean property, we can find integer $n$ such that: $$3^n > 2n > \frac{1}{b - a} \ \Rightarrow \ \frac{2}{3^n} < b - a$$