I was reading a book and I came across the following exercise: "Let $M$ be a compact metric space and let $f:M\longrightarrow M$ satisfy $d(f(x),f(y))\geq d(x,y)$ for all $x,y\in M$. Prove that $f$ is an isometry of M onto itself."
The hint in the book suggests that for a fixed $x\in M$, we should consider the sequence $x_{n}=f^{n}(x)$ and prove, possibly by passing to a subsequence, that the sequence, that the sequence $\{x_{n}\}_{n=1}^{\infty}$ converges to $x$.
I am having trouble proving this part of the hint. Since $M$ is compact, the sequence $\{f^{n}(x)\}_{n=1}^{\infty}$ has a subsequence that converges. Thus, it is a Cauchy sequence.
Using the definition of a Cauchy sequence, I have concluded that for any $\varepsilon>0$, there exists $N\in\mathbb{Z}^{+}$ such that if $j>i\geq N$ then $$d(x,f^{n_{I}-n_{j}}(x))\leq d(f^{n_{j}}(x),f^{n_{i}}(x))<\varepsilon.$$
However, I am not sure how to conclude that the entire sequence $\{f^{n}(x)\}_{n=1}^{\infty}$ converges to $x$ because for convergence, this inequality should hold for every $n-th$ iteration of $f$, not just for the indices where the Cauchy condition holds.
Can anyone provide guidance on how to proceed from here?
