Can this intuition give a proof that an isometry $f:X \to X$ is surjective for compact metric space $X$? A prelim problem asked to prove that if $X$ is a compact metric space, and $f:X \to X$ is an isometry (distance-preserving map) then $f$ is surjective. The official proof given used sequences/convergent subsequences and didn't appeal to my intuition. When I saw the problem, my immediate instinct was that an isometry should be "volume-preserving" as well, so the volume of $f(X)$ should be equal to the volume of $X$, which should mean surjectivity if $X$ is compact. The notion of "volume" I came up with was the minimum number of $\epsilon$-balls needed to cover $X$ for given $\epsilon > 0$. If $f$ were not surjective, then because $f$ is continuous this means there must be a point $y \in X$ and $\delta > 0$ so that the ball $B_\delta(y)$ is disjoint from $f(X)$. I wanted to choose $\epsilon$ in terms of $\delta$ and use the fact that an isometry carries $\epsilon$-balls to $\epsilon$-balls, and show that given a minimum-size cover of $X$ with $\epsilon$ balls, that a cover of $X$ with $\epsilon$-balls could be found with one fewer ball if $f(X) \cap B_\delta(y) = \emptyset$, giving a contradiction. Can someone see a way to make this intuition work?

That's a nice idea for a proof. I think perhaps it works well to turn it inside out, so to speak:
Lemma. Assuming $X$ is a compact metric space, for each $\delta>0$ there is a finite upper bound to the number of points in $X$ with a pairwise distance $\ge\delta$. (Let us call such a set of points $\delta$-separated.)
Proof. $X$ is totally bounded, so there exists a finite set $N$ of points in $X$ so that every $x\in X$ is closer than $\delta/2$ to some member of $N$. Any two points in $B_{\delta/2}(x)$ are closer together than $\delta$, so there cannot be a $\delta$-separated set with more members than $N$.
Now let $f\colon X\to X$ be an isometry and not onto. Let $x\in X\setminus f(X)$, and let $\delta>0$ be the distance from $x$ to $f(X)$. Let $E\subseteq X$ be a $\delta$-separated set with the largest possible number of members. Then $f(E)\cup\{x\}$ is such a set with more members. Contradiction.
