TITLE: A metric space such that all closed balls are compact is complete.
QUESTION [13 upvotes]: I am trying to solve the following exercise:

Let $(X,d)$ be a metric space that has the property that for any $x\in X$ and $r>0$, the closed ball
  $$\bar{B}(x,r):=\{y\in X:d(x,y)\leq r\}$$
  is compact. Show that $X$ is complete.

I think I have a proof, but I am only using that the unit closed balls $\bar{B}(x,1)$ are compact. Maybe I am missing something?
Attempt: Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $(X,d)$. Then, $\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(x_n,x_N)<1$. Thus, $\forall n\geq N$, $x_n\in\bar{B}(x_N,1)$ which is compact by assumption, so $(x_n)_{n=N}^\infty$ is a sequence in this compact set and thus has a convergent subsequence. Hence, $(x_n)_{n=1}^\infty$ is a Cauchy sequence that has a convergent subsequence, so it converges in $(X,d)$. Therefore, $(X,d)$ is complete.

REPLY [4 votes]: If there is an $r>0$ with $\overline{B(x,r)}$ compact for all $x$, then for any $r>\varepsilon>0$, $\overline{B(x,\varepsilon)}$ is compact. It's not so much stronger that you're just using compactness of $B(x,1)$ for all $x$; that implies compactness at all smaller scales.