Let M be a compact metric space and $f:M\rightarrow M$ be such that $f$ is continuous and $d(f(x),f(y))\geq d(x,y)$, for all $x,y \in M$. Show that, for all $x$ in $M$, there exists $n_1<n_2<...$ such that $\lim_{k\rightarrow \infty}f^{n_k}(x)=x$.
I've been trying to show the converse implies $f^{n}(x)$ has no convergent subsequence, but it's not working out...
Here is my latest attempt at it, if someone could tell me if my reasoning is correct I would appreciate it: Suppose that, for all $n$ $d(f^n(x),x)\geq1$. Then, for all $m>n$: $$d(f^m(x),f^n(x))\geq ... \geq d(f^{m-n}(x),x)\geq1$$ Thus the sequence $\{f^n(x)\}_n$ has no convergent subsequence, absurd since $M$ is compact. So there exists $n_1$ such that $d(f^{n_1}(x),x)<1$ If there exists $n_2\leq n_1$ such that $d(f^{n_2}(x),x)<1/2$, redefine $n_1$ so it's $n_2$.
-This is the part I'm a bit insecure: So we may assume that $\forall n\leq n_1,\, d(f^n(x),x)\geq 1/k_1$, for some $k_1>1$.
Therefore, if we suppose that for all $n>n_1$, $d(f^n(x),x)\geq 1/k_1$, then by the same argument as before, the sequence would have no convergent subsequence, thus there exists $n_2>n_1$ such that $d(f^{n_2}(x),x)<1/k_1\leq1/2$. By the same argument as before, we may assume $d(f^{n_2}(x),x)\geq1/k_2$, with $k_2>k_1$.
Then we proceed inductively, obtaining a the subsequence $f^{n_k}(x)$ such that $d(f^{n_k}(x),x)<1/k$, thus $f^{n_k}(x)\rightarrow x$.