Prob. 7 (b), Sec. 28, in Munkres' TOPOLOGY, 2nd ed: A shrinking self-map of a compact metric space has a unique fixed point

Solution 1:

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.