A metric space such that all closed balls are compact is complete.

Your proof is right.

As a justification that it is not strange that you needed a little less than the hypothesis, let me show you what happens when your metric space is a normed space: since any two balls are homeomorphic, if one closed ball is compact than all are. Indeed, $$ \bar B(0,1)\simeq \bar B(y,r) $$ via the continuous function $x\mapsto r(x+y)$.

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.