I'm trying to prove a result mentioned in this thread, i.e.,
Let $(X, d)$ be a compact metric space and $f:X \to X$ an isometry. Then $f$ is surjective.
Could you confirm if my below attempt is fine?
Proof Assume the contrary that there is $a \in X$ such that $a \notin f(X)$. Because $f$ is continuous and $X$ compact, we get $f(X)$ is compact. So there is $r>0$ such that $$ d(a, f(X)) := \inf_{x \in X} d(a, f(x)) \ge r. $$
We define $(x_n, n \in \mathbb N)$ by $x_0 := a$ and $x_{n+1} := f(x_n)$. Because $f$ is an isometry, $$ d(x_n, x_m) = d(x_0, x_{m-n}) \ge d(a, f(X))\ge r \quad \forall m > n. $$
Because $X$ is compact, there is a sub-sequence $\varphi$ of $\mathbb N$ and $b \in X$ such that $x_{\varphi (n)} \to b$. This is a contradiction because $$ d(x_{\varphi (n)}, x_{\varphi (m)}) \ge r \quad \forall m \neq n. $$
This completes the proof.
