This is without a doubt the best problem I've encountered in Munkres so far! This proof is multi-stepped, and I promise that each step
is more enthralling than the last! We are given that $d$ is a metric for $X$, and we must show that:
$$d'(x, \ y) \ = \ \frac{d(x, \ y)}{1 \ + \ d(x, \ y)}$$
Is a bounded metric that gives the topology of $X$. Munkres gives us a hint! He tells us to use mean-value theorem in order to demonstrate that
$f(a \ + \ b) \ - \ f(b) \ \leq \ f(a)$, where:
$$f(x) \ = \ \frac{x}{1 \ + \ x}$$
Mean-value theorem states that for some continuous function on the interval $[a, \ b]$, and differentiable on the interval $(a, \ b)$, there exists some
$c \ \in \ (a, \ b)$ such that:
$$\frac{f(b) \ - \ f(a)}{b \ - \ a} \ = \ f'(c)$$
This is basically saying that there exists at least one point on any interval on a continuous function such that the derivative of the function evaluated at
that point is the same as the slope evaluated by finding the secant line through the start and end points of the interval, lying on the curve. Ok, fun. Now,
Let's pick an interval of the curve $f$ of $(a, \ a \ + \ b)$. We then have:
$$\frac{f(a \ + \ b) \ - \ f(a)}{b} \ = \ f'(c) \ \Rightarrow \ f(a \ + \ b) \ - \ bf'(c) \ = \ f(a)$$
So we have to prove that $f(a \ + \ b) \ - \ f(b) \ \leq \ f(a \ + \ b) \ - \ bf'(c) \ \Rightarrow \ f(b) \ \geq \ bf'(c) \ \Rightarrow \ \frac{f(b)}{b} \ \geq \ f'(c)$.
Let's take the derivative of $f$:
$$f'(x) \ = \ \frac{\text{d}}{\text{dx}} \frac{x}{1 \ + \ x} \ = \ \frac{1}{1 \ + \ x} \ - \ \frac{x}{(1 \ + \ x)^2} \ = \ \frac{1}{(1 \ + \ x)^2} \ \Rightarrow \ f'(c) \ = \ \frac{1}{(1 \ + \ c)^2}$$
And on the left side of the inequality:
$$\frac{f(b)}{b} \ = \ \frac{1}{1 \ + \ b}$$
Now, we assert that $\frac{1}{1 \ + \ b} \ > \ \frac{1}{(1 \ + \ c)^2}$ when $b \ < \ c$ and $b, \ c \ > \ 0$, which is obviously true, therefore we have demonstrated that $f(a \ + \ b) \ - \ f(b) \ \leq \ f(a)$
(we have shown that at point $c$, we can prove this condition, but since $b$ remains fixed, it holds true no matter the chosen value of $c$).
Next, we have to show that $d'$ is a bounded metric. The first two conditions of the metric are trivial, as we are given $d(x, \ y)$ is a valid metric. To prove the triangle inequality, we must show that:
$$d'(x, \ z) \ \leq \ d'(x, \ y) \ + \ d'(y, \ z) \ \Rightarrow \ \frac{d(x, \ z)}{1 \ + \ d(x, \ z)} \ \leq \ \frac{d(x, \ y)}{1 \ + \ d(x, \ y)} \ + \ \frac{d(y, \ z)}{1 \ + \ d(y, \ z)}$$
Remember, we just proved that $f(a \ + \ b) \ - \ f(b) \ \leq \ f(a) \ \Rightarrow \ f(a \ + \ b) \ \leq \ f(a) \ + \ f(b)$. Let $a \ = \ d(x, \ y)$ and $b \ = \ d(y, \ z)$. We have:
$$\frac{d(x, \ y) \ + \ d(y, \ z)}{1 \ + \ d(x, \ y) \ + \ d(y, \ z)} \ \leq \ \frac{d(x, \ y)}{1 \ + \ d(x, \ y)} \ + \ \frac{d(y, \ z)}{1 \ + \ d(y, \ z)}$$
Now, observe that the function $f$ (and therefore $d'$) is strictly increasing (look at the derivative and ask yourself, can this function ever be negative? The correct answer is no). Recalling that $d$ is a metric, we have:
$$d(x, \ z) \ \leq \ d(x, \ y) \ + \ d(y, \ z)$$
Because of this, and the fact that the function is strictly increasing, we have:
$$\frac{d(x, \ z)}{1 \ + \ d(x, \ z)} \ \leq \ \frac{d(x, \ y) \ + \ d(y, \ z)}{1 \ + \ d(x, \ y) \ + \ d(y, \ z)} \ \leq \ \frac{d(x, \ y)}{1 \ + \ d(x, \ y)} \ + \ \frac{d(y, \ z)}{1 \ + \ d(y, \ z)}$$
And so we have proved that $d'$ is a valid metric.
This metric is clearly bounded, as:
$$\displaystyle\lim_{d(x, \ y) \ \rightarrow \infty} \ \frac{d(x, \ y)}{1 \ + \ d(x, \ y)} \ = \ 1$$
Now, all that's left is to prove that it gives the topology for $X$. I'm assuming that $d$ gives the topology on $X$ that we are trying to find, therefore we have to prove that
the topologies generated by $d$ and $d'$ are equivalent. To do this, we simply have to prove that for each $x$ in $X$ and each $\epsilon$, we have $\delta$ values such that:
$$B_d(x, \ \delta) \ \subset \ B_{d'}(x, \ \epsilon) \ \ \ \ \text{and} \ \ \ \ B_{d'}(x, \ \delta) \ \subset \ B_{d}(x, \ \epsilon)$$
To prove the first condition, let us pick $B_{d'}(x, \ \epsilon)$. We must show that there is some $B_d(x, \ \delta)$ contained within $B_{d'}(x, \ \epsilon)$. By definition, we know that for any $y \ \in \ B_{d'}(x, \ \epsilon)$,
it must be true that:
$$d'(x, \ y) \ = \ \frac{d(x, \ y)}{1 \ + \ d(x, \ y)} \ < \ \epsilon \ \Rightarrow \ d(x, \ y) \ < \ \epsilon \ + \ \epsilon d(x, \ y)) \ \Rightarrow \ (1 \ - \ \epsilon)d(x, \ y) \ < \ \epsilon \ \Rightarrow \ d(x, \ y) \ < \ \frac{\epsilon}{1 \ - \ \epsilon}$$
So if we chose $\delta \ = \ \frac{\epsilon}{1 \ - \ \epsilon}$, then for any $y \ \in \ B_d(x, \ \delta)$, $y$ will also be in $B_{d'}(x, \ \epsilon)$, meaning that $B_d(x, \ \delta) \ \subset \ B_{d'}(x, \ \epsilon)$. On the other hand, say we have some $y \ \in \ B_{d}(x, \ \epsilon)$, then it is true that:
$$d(x, \ y) \ < \ \epsilon$$
We need to show that there is some open ball in the metric $d'$ that is contained within the open ball in $d$. Let us pick $y$ to be in the open ball in $d'$. We then have:
$$\frac{d(x, \ y)}{1 \ + \ d(x, \ y)} \ < \ \delta \ \Rightarrow \ d(x, \ y) \ < \ \delta \ + \ \delta d(x, \ y) \ \Rightarrow \ d(x, \ y) \ < \ \frac{\delta}{1 \ - \ \delta}$$
But we require that $d(x, \ y) \ < \ \epsilon$, so we set up:
$$\frac{\delta}{1 \ - \ \delta} \ = \ \epsilon \ \Rightarrow \ \delta \ = \ \epsilon \ - \ \epsilon \delta \ \Rightarrow \ \delta(1 \ + \ \epsilon) \ = \ \epsilon \ \Rightarrow \ \delta \ = \ \frac{\epsilon}{1 \ + \ \epsilon}$$
So in this case, if we chose $\delta \ = \ \frac{\epsilon}{1 \ + \ \epsilon}$, then for any $y \ \in \ B_{d'}(x, \ \delta)$, then $y \ \in \ B_{d}(x, \ \epsilon)$, therefore showing that $B_{d'}(x, \ \delta) \ \subset \ B_{d}(x, \ \epsilon)$.
We have therefore proven that the two metrics generate the same topology on $X$, and therefore that $d'$ gives the topology of $X$, as was stated in the original question.