Locally compact metric space 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 ?
 A: 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.
A: 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.
A: A metric space is proper iff (by definition) all closed balls are compact.
https://en.wikipedia.org/wiki/Metric_space#Locally_compact_and_proper_spaces
All proper spaces are locally compact (see the link above) (hence "if" is true in your claim) and complete (see the link below):
https://en.wikipedia.org/wiki/Glossary_of_Riemannian_and_metric_geometry#P
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.
