Prob. 7 (b), Sec. 28, in Munkres' TOPOLOGY, 2nd ed: A shrinking self-map of a compact metric space has a unique fixed point Here is Prob. 7, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:

Let $(X, d)$ be a metric space. If $f$ satisfies the condition
  $$ d\big( f(x), f(y) \big) < d(x, y) $$
  for all $x, y \in X$ with $x \neq y$, then $f$ is called a shrinking map. If there is a number $\alpha < 1$ such that
  $$ d \big( f(x), f(y) \big) \leq \alpha d(x, y) $$
  for all $x, y \in X$, then $f$ is called a contraction. A fixed point of $f$ is a point $x$ such that $f(x) = x$. 
(a) If $f$ is a contraction and $X$ is compact, show $f$ has a unique fixed point. [Hint: Define $f^1 = f$ and $f^{n+1} = f \circ f^n$. Consider the intersection $A$ of the sets $A_n = f^n(X)$.] 
(b) Show more generally that if $f$ is a shrinking map and $X$ is compact, then $f$ has a unique fixed point. [Hint: Let $A$ be as before. Given $x \in A$, choose $x_n$ so that $x = f^{n+1}\left(x_n\right)$. If $a$ is the limit of some subsequence of the sequence $y_n = f^n \left( x_n \right)$, show that $a \in A$ and $f(a) = x$. Conclude that $A = f(A)$, so that $\mathrm{diam}\, A = 0$.]
(c) Let $X = [0, 1]$. Show that $f(x) = x - x^2/2$ maps $X$ into $X$ and is a shrinking map that is not a contraction. [Hint: Use the mean-value theorem of calculus.]
(d) The result in (a) holds if $X$ is a complete metric space, such as $\mathbb{R}$; see the exercises of \Sec. 43. The result in (b) does not: Show that the map $f \colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = \left[ x + \left( x^2 + 1 \right)^{1/2} \right]/2$ is a shrinking map that is not a contraction and has no fixed point.

Here is my MSE post on Prob. 7 (a).
Here I'll only be attempting a solution to Prob. 7 (b).
My Attempt 
Prob. 7 (b):
Here is another Math SE post on this very problem. However, here I'll attempt a proof using the hint offered by Munkres.

We first show that the shrinking map $f$ is uniformly continuous on $X$. Given a real number $\varepsilon > 0$, let us choose a real number $\delta$ so that $0 < \delta \leq \varepsilon$. Then for all $x, y \in X$ for which $d(x, y) < \delta$, we would obtain
  $$ d \big( f(x), f(y) \big) \leq d( x, y) < \delta \leq \varepsilon.$$
  Since $\varepsilon > 0$ was arbitrary, it follows that $f$ is uniformly continuous on $X$.
Let $i_X \colon X \to X$ denote the identity map on $X$, defined by 
  $$ i_X (x) \colon= x \ \mbox{ for all } \ x \in X. \tag{Def. 0} $$
  Now let us put 
  $$ f^n \colon= \begin{cases} i_X \ & \mbox{ if } n = 0, \\ f \circ f^{n-1} \ & \mbox{ if } n = 1, 2, 3, \ldots. \end{cases} \tag{Def. 1} $$
  Next, let us put
  $$ A_n \colon= \begin{cases} X \ & \mbox{ if } n = 0, \\ f^n(X) \ & \mbox{ if } n = 1, 2, 3, \ldots. \end{cases} \tag{Def. 2} $$
  Then we find that, for each natural number $n$, 
  $$ A_n = f \left( A_{n-1} \right). \tag{0} $$
Now as the maps $i_X$ and $f$ are both continuous mappings of the compact space $X$ into itself, so are all the maps $f^n$ in (Def. 1) above, and thus all the sets $A_n$ in (Def. 2) above are all compact subspaces of $X$; moreover since $X$, being a metric space, is a Hausdorff space and since each set $A_n$ is a compact subspace of $X$, each set $A_n$ is also closed in $X$. And, as each set $A_n$ is closed in $X$, so is the intersection of these sets.
  Let us put $$ A \colon= \bigcap_{n=0}^\infty A_n. \tag{Def. 3}$$
  Then as $A$ is a closed set in the compact space $X$, so $A$ is also compact (as a subspace of $X$).
As $f$ is a mapping of set $X$ into itself, so we have $f(X) \subset X$, that is, $$A_1 \subset A_0.$$
  Now suppose that, for some natural number $k$, we have $$ A_k \subset A_{k-1}. $$
  Then using (0) above we find that
  $$ A_{k+1} = f \left( A_k \right) \subset f \left( A_{k-1} \right) = A_k.$$
  Therefore by induction we can conclude that 
  $$ A_n \subset A_{n-1} \ \mbox{ for } n = 1, 2, 3, \ldots. \tag{1} $$
Thus $\left\{ \ A_n \ \colon \ n = 0, 1, 2, \ldots \ \right\}$ is a nested sequence of non-empty closed sets in the compact space $X$; therefore their intersection is non-empty, that is, set $A$ in (Def. 3) above is non-empty.
We now show that $\mathrm{diam}\, X$ is finite. Let $p$ be any point of $X$. Then the collection 
  $$ \left\{ \ B_d \left(p, N \right) \ \colon \ N \in \mathbb{N} \ \right\},$$
  where 
  $$ B_d \left( p; N \right) \colon= \{ \ x \in X \ \colon \ d(x, p) < N \ \},$$
  forms an open covering of the compact space $X$; so some finite sub-collection of this collection also covers $X$; that is, there exist finitely many natural numbers $N_1, \ldots,  N_n$ such that the collection 
  $$ \left\{ \ B_d \left(p, N_1 \right), \ldots, B_d \left(p, N_n \right) \ \right\}$$
  of open balls covers $X$. Let 
  $$ M \colon= \max\left\{ \ N_1, \ldots, N_n \ \right\}. $$
  Then we obtain
  $$ X = B_d (p, M).$$
  Thus for any points $x, y \in X$, we have
  $$ d(x, y) \leq d(x, p) + d(p, y) < M + M = 2M.$$
  So $$ \mathrm{diam}\, X \leq 2M < +\infty. $$
  Thus we have shown that 
  $$ \mathrm{diam}\, X < +\infty. \tag{2} $$
  Hence from (Def. 3) above and from (1) we can also conclude that 
  $$ \mathrm{diam}\, A \leq  \mathrm{diam}\, A_n \leq \mathrm{diam}\, A_{n-1} < +\infty \ \mbox{ for } n = 1, 2, 3, \ldots. \tag{3} $$
Now suppose $x \in A$. Then $x$ is in each set $A_n = f^n(X)$, and so there exists a point $x_n \in X$ such that $x = f^{n+1}\left(x_n\right)$ for each $n = 1, 2, 3, \ldots$; let us put
  $$ y_n \colon= f^n\left(x_n\right) \ \mbox{ for each } n = 1, 2, 3, \ldots. \tag{Def. 4} $$
  Then $\left( y_n \right)_{n \in \mathbb{N}}$ being a sequence in the compact metric space $(X, d)$ has a convergent subsequence, by virtue of Theorem 28.2 in Munkres; let $\left( y_{\varphi(n)} \right)_{n \in \mathbb{N}}$ be this subsequence for some strictly increasing function $\varphi \colon \mathbb{N} \to \mathbb{N}$, and let $a$ be the limit of this sequence.

What next? How to proceed from here?
PS:

Thus, for each $n \in \mathbb{N}$, we have 
  $$ y_n = f^n \left( x_n \right)$$
  and also
  $$ x = f^{n+1} \left( x_n \right) = f \left( f^n \left( x_n \right) \right) = f\left( y_n \right). $$
  and so 
  $$ f \left( y_n \right) = x. $$
  Therefore we can conclude that, for each $n \in \mathbb{N}$, we have
  $$ f \left( y_{\varphi(n)} \right) = x, $$
  and therefore we obtain
  $$ \lim_{n \to \infty} f \left( y_{\varphi(n)} \right) = x. \tag{4*} $$
  But as $f$ is continuous and as 
  $$ \lim_{n \to \infty} y_{\varphi(n)} = a, $$
  so we must also have 
  $$ \lim_{n \to \infty} f \left( y_{\varphi(n)} \right) = f(a). \tag{4**} $$
  But the limit of a sequence in any metric space is unique. Therefore from (4*) and (4**) we obtain 
  $$ f(a) = x, \tag{4}  $$
  and as $x \in A$ and $a \in X$, so from (4) we obtain $x \in f(A)$, 
  which implies that 
  $$ A \subset f(A). \tag{5*} $$
On the other hand, if $p \in f(A)$, then we have $p = f(q)$ for some point $q \in A$. But as 
  $$ A = \cap_{n = 0}^\infty A_n, $$
  and as $q \in A$, so $q \in A_n$ and hence  $p = f(q) \in f \left( A_n \right) = A_{n+1}$ for each $n = 0, 1, 2, 3, \ldots$. That is, $p \in A_n$ for each $n = 1, 2, 3, \ldots$. But $p \in X = A_0$ of course. Therefore we can conclude that $p \in \cap_{n=0}^\infty A_n = A$, from which it follows that 
  $$ f(A) \subset A. \tag{5**} $$
  From (5*) and (5**) we obtain 
  $$ f(A) = A. \tag{5} $$
We note that 
  $$ \mathrm{diam}\, A = \sup \big\{ \, d(x, y) \, \colon \, x, y\in A \, \big\}. \tag{Def. 4} $$
  Moreover from  (3) we obtain 
  $$ 0 \leq \mathrm{diam}\, A < +\infty. $$
Now suppose that $\mathrm{diam}\, A > 0$. Then in view of (Def. 4) we can conclude that, for any real number $\varepsilon > 0$, we can find points $x_\varepsilon, y_\varepsilon \in A$ such that 
  $$ \mathrm{diam}\, A \geq d \left( x_\varepsilon, y_\varepsilon \right) > \mathrm{diam}\, A  - \varepsilon. $$
  So for each $n \in \mathbb{N}$, we can find points $a_n, b_n \in A$ such that 
  $$ \mathrm{diam}\, A \geq d \left( a_n, b_n \right) > \mathrm{diam}\, A - \frac1n. \tag{6} $$ 
  Thus we obtain sequences $\left( a_n \right)_{n \in \mathbb{N}}$ and $\left( b_n \right)_{n \in \mathbb{N}}$ in set $A$. 
Now as $A$ is compact and as $\left( a_n \right)_{n \in \mathbb{N}}$ is a sequence in $A$, so there exists a subsequence $\left( a_{\phi(n)} \right)_{n \in \mathbb{N}}$ of this sequence, where $\phi \colon \mathbb{N} \to \mathbb{N}$ is a strictly increasing function, such that $\left( a_{\phi(n)} \right)_{n \in \mathbb{N}}$ converges to some point $a$ in $A$.
And, as $\left( b_{\phi(n)} \right)_{n \in \mathbb{N}}$ is a sequence in $A$ and as $A$ is compact, so there exists a subsequence $\left( b_{\psi(n)} \right)_{n \in \mathbb{N}}$ of this sequence that converges to some point $b$ in $A$. Here $\psi \colon \mathbb{N} \to \mathbb{N}$ is a strictly increasing function such that 
  $$ \mathrm{range}\, \psi \subset \mathrm{range}\, \phi. $$
  Then $\left( a_{\psi(n)} \right)_{n \in \mathbb{N} }$ is a subsequence of $\left( a_{\phi(n)} \right)_{n \in \mathbb{N} }$, and as the latter converges to point $a \in A$, so the former also converges to $a$. 
Now as 
  $$ \lim_{n \to \infty} a_{\psi(n)} = a \ \mbox{ and } \ \lim_{n \to \infty} b_{\psi(n)} = b, $$
  so we can conclude that 
  $$ \lim_{n \to \infty} d \left( a_{\psi(n)}, b_{\psi(n)} \right) = d(a, b).  $$
  But from (6) we obtain
  $$ \mathrm{diam}\, A \geq d \left( a_{\psi(n)}, b_{\psi(n)} \right) > \mathrm{diam}\, A - \frac{1}{\psi(n)} \geq \mathrm{diam}\, A - \frac{1}{n}, $$
  and upon taking the limit as $n \to \infty$ we obtain 
  $$ \mathrm{diam}\, A \geq d(a, b) \geq \mathrm{diam}\, A, $$
  and so
  $$ d(a, b) = \mathrm{diam}\, A > 0 \tag{7} $$

Another way to show that (7) holds is as follows: 

Suppose $\mathrm{diam} \, A > 0$. 
As $A$ is compact, so too is the Cartesian product $A \times A$, by Theorem 26.7 in Munkres.
As the metric function $d \colon X \to X \to \mathbb{R}$ is a continuous map, so too is the restriction $d|_{A\times A}$, and as $A \times A$ is compact, so the map $d|_{A\times A}$ has a largest element, namely $\mathrm{diam}\, A$ of course. That is, there exist points $a, b \in A$ such that 
  $$ d(a, b) = \mathrm{\diam}\, A > 0. \tag{7} $$
  Therefore 
  $$ a \neq b. \tag{8} . $$
However, as $A = f(A)$ by (5) above and as $a, b \in A$, so we can conclude that $a, b \in f(A)$ also, which implies that 
  $$ a = f\left( a^* \right) \ \mbox{ and } \ b = f\left( b^* \right) $$
  for some points $a^*, b^* \in A$, and as $a \neq b$ by (8) above, so we also have $a^* \neq b^*$, and since $f$ is a shrinking map, this together with (7) above implies that
  $$  \mathrm{diam}\, A = d(a, b) = d \big( f \left( a^* \right), f \left( b^* \right) \big) < d \left( a^*, b^* \right). $$
  But on the other hand, as $a^*, b^* \in A$, so we must have 
  $$ d \left( a^*, b^* \right) \leq \mathrm{diam}\, A. $$
  Thus we have a contradiction. Therefore our supposition that $\mathrm{diam}\, A > 0$ is wrong. Hence 
  $$ \mathrm{diam}\, A = 0. \tag{9} $$
  Thus $A$ has only one point in it; let $p$ be that point. Then as $p \in A$, so $f(p) \in f(A)$. But by (5) we have $f(A) = A$. So we must have $f(p) \in A$ also, and since $A$ has only one element $p$, therefore we can conclude that 
  $$ f(p) = p. $$
  Thus $f$ has a fixed point $p \in A \subset X$.
Finally if $p$ and $q$ were any two distinct fixed points of the shrinking map $f$, then we would obtain 
  $$ d(p, q) = d \big( f(p), f(q) \big) < d(p, q), $$
  a contradiction. Hence $f$ has a unique fixed point in $X$. 

Is my proof correct now? Is it complete and clear in all respects? Or, are there any gaps in it of logic or clarity?
 A: We have $y_{\varphi(n)} \in A_{\varphi(n)} \subset A_{\varphi(m)}$ for $n \ge m$. Hence $a = \lim y_{\varphi(n)} \in A_{\varphi(m)}$ because $A_{\varphi(m)}$ is closed. This implies that $a \in \bigcap_m A_{\varphi(m)} = A$. Since $f$ is continuous and $y_{\varphi(n)} \to a$, we get $f(y_{\varphi(n)}) \to f(a)$. But the sequence $f(y_{\varphi(n)}) = f^{\varphi(n)+1}(x_{\varphi(n)}) = x$ is constant and we conclude $f(a) = x$.
This shows $A \subset f(A)$.
Assume that $d = \text{diam} A > 0$. Then we find sequences $(x_n), (y_n)$ in $A$ such that $d(x_n,y_n) \to d$. Since $A$ is compact, we may  w.l.o.g. assume that both sequence converge to points $x, y \in A$. We get $d(x,y) = d$. Choose $a, b \in A$ such that $f(a) = x, f(b) = y$. Then $d = d(x,y) = d(f(a),f(b)) < d(a,b)$, which contradicts the definition of $d$.
Therefore $\text{diam} A = 0$ which is possible only when $A$ contains a single point $a$. This is a fixed point of $f$. Since $A$ trivially contains all fixed points of $f$, we are done.
A: To show that diam$A=0$, we only need to note that since $A\times A$ is compact, we must have a point $(x,y)$ s.t.$d(x,y)=diam A$ by extreme value theorem.
