Prove that if in a metric space all closed balls are compact, a subset is compact if and only if it is closed and bounded.
Attempt: If all closed balls are compact, then there is a converging subsequence, so the subset is closed and bounded since the subsequence converges. Is that right? How do I show the other direction?
Please help and thank you!
Suppose $C$ is closed and bounded set in $X$, then due to boundedness $\exists R>0$ such that $C\subset [-R,R]$. But $[-R,R]$ is compact as per hypothesis and any closed subset of a compact space is itself compact.