I'm trying to prove that a metric space is locally compact iff every closed ball is compact, using the more general definition that applies to Hausdorff spaces, that every point has a compact neighbourhood. So call $X$ my space. The only non trivial thing to prove is that every closed ball is compact, assuming $X$ is locally compact. So consider $N$ a compact neighbourhood of some $x\in X$. Then as a neighbourhood, it contains $B(x,r)$ for some $r$. So it contains $\bar{B}(x,r/2)$. This is closed inside $N$ which is compact, so it's also compact. So I've proven that at any point there is a compact closed neighbourhood ball. Surely it's not too hard to prove all the bigger closed balls are compact ?
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$\begingroup$Consider $R^2-\{(0,0)\}$ endowed with the canonical metric, it is locally compact. But $B((0,1);2)$ is not compact.
But the result is true if $X$ is endowed with a norm.
$\endgroup$ 1 $\begingroup$If $(X,d)$ is a metric space, then $d'(x,y) = d(x,y)/(1+d(x,y))$ defines a new metric having the same (open or closed) balls as $(X,d)$. But then $X$ is the ball of radius $1$ centered anywhere. Hence if, in addition, $X$ is locally compact but not compact, the metric space $(X,d')$ admits a closed non compact ball.
$\endgroup$ $\begingroup$A metric space is proper iff (by definition) all closed balls are compact.
All proper spaces are locally compact (see the link above) (hence "if" is true in your claim) and complete (see the link below):
So any incomplete locally compact metric space is a counter-example to "only if".
Moreover, as mentioned Tsemo Aristide's answer, any non-compact metric space, even a proper one, has the same topology as some improper metric space.
A normed space X is proper iff it is locally bounded (iff it is finite-dimensional).
This last claim follows from Theorem 1.22 of Rudin: Functional Analysis. That says that locally compact topological vector spaces are finite-dimensional, hence equivalent to some $\mathbb K^n$, by Theorem 1.21, where $\mathbb K$ is your scalar field (i.e., $\mathbb R$ or $\mathbb C$). $\mathbb K^n$ being proper is well known.
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