This question originates from this terrific Answer by Qiaochu Yuan. For convenience, I copy it here.
Proposition: Any isometry[isometric mapping] $f : X \to X$ of a compact metric space is bijective.
Proof. $f$ is clearly injective. Suppose $f$ is not bijective. Then $f(X)$ is compact, so given $x \in X \setminus f(X)$ the distance $\text{dist}(x, f(X))$ is positive. Pick $\epsilon < \text{dist}(x, > f(X))$. Let $N$ be the smallest positive integer for which $X$ admits a cover by $N$ open sets of diameter less than $\epsilon$. No such set containing $x$ can intersect $f(X)$, but by pulling back along $f$ it follows that we can find a cover of $X$ by $N-1$ open sets of diameter less than $\epsilon$; contradiction.
The answer assumes that there exist $x$ such that $\text{dist}(x, f(X))$ is positive, that is to say, $X\backslash f(X)$ has an non-empty interior. This assumption is true due to the fact that $X$ is compact and thus complete.
However, the rest of the proof does not rely on the fact that $X$ is compact. It's true that compactness implies the existence of a cover by $N<\infty$ open sets of diameter less than $\epsilon$, but the latter does not imply compactness. To ensure the existence of such a cover, we just need $X$ to be totally bounded, i.e. $\forall \epsilon,\exists a_1,\ldots,a_n\in X, X\subseteq \cup_{i=1}^nB(a_i,\epsilon)$.
So can anyone give a (counter)example, showing that if $X$ is not compact but is totally bounded, then $f$ can FAIL to be surjective?
