Let $M$ be a compact metric space, and suppose $f:M \to M$ is short, i.e
$$ d(f(x),f(y)) \le d(x,y) \, \,\text{ for all } \, x,y \in M.$$
Define $X=\bigcap_{n=1}^{\infty}\, f^n(M)$.
Question: How to prove that $f|_X:X \to X$ is an isometry?
I know that any surjective short map from a compact metric space into itself is an isometry, and $X$ is compact. Is $f|_X$ surjective?
Yes, $f |_X$ is surjective.
To see why, pick $x \in X$, and so $x \in f^{n+1}(M)$ for all $n \ge 0$. We may therefore pick $y_n \in f^n(M)$ such that $f(y_n)=x$. Since $M$ is a compact metric space, the sequence $y_n$ has a subsequence $y_{n(i)}$ converging to a limit $y$. It follows that $f(y)=x$.
If $y$ were not in $X$ then, since $X$ is closed and $M$ is normal, we could find disjoint open sets $U,V \subset M$ such that $y \in U$ and $X \subset V$. For sufficiently large $n$ we would then have $f^n(M) \subset V$, hence for sufficiently large $i$ we would have $y_{n(i)} \in U \cap V$, a contradiction.
Added: The reason that $f^n(M) \subset V$ for sufficiently large $n$ is that $f^n(M) \cap (M-V)$ is a nested sequence of compact sets, so if it were not eventually empty then its intersection would be nonempty, contradicting that the nested intersection of the sequence $f^n(M)$ equals $X \subset V$.