# Locally Euclidean Hausdorff topological space is topological manifold iff $\sigma$-compact.

I'd like somebody to specify flaws in my outline of the proof of the above statement. I'm following the definition of topological manifold used in Lee's Introduction to Smooth Manifolds. (it is 2nd countable)

Let $X$ to be a Locally Euclidean Hausdorff topological space.

$\Rightarrow$: If $X$ is a topological manifold, then it has a countable basis of precompact coordinate balls by a lemma. Therefore, there is a countable set of precompact coordinate balls which cover $X$, and the set of closure of these balls is the set of countably many compact subspaces which cover $X$. Thus, $X$ is $\sigma$-compact.

$\Leftarrow$: If $X$ is $\sigma$-compact, there is a set of countably many compact sets which cover $X$. Let $C$ be an arbitrary compact set of this set. Since $X$ (and $C$) is locally Euclidean, $C$ is locally metrizable. Since $C$ is locally metrizable compact Hausdorff space, it is metrizable and therefore 2nd countable. Thus, being the countable union of these compact spaces with countable basis, $X$ is 2nd countable and therefore a topological manifold.

Should I add a proof that locally metrizable compact Hausdorff space is metrizable? I found this statement in Munkre's Topology, but is it really a widely known fact?

The manifold $\Rightarrow \sigma-$compact looks fine. The other direction probably also works, but you can prove it much more directly/easier.
It suffices to show $X$ the space is second countable. Let $K_n$ be the covering of $X$ by compact sets, and assume they're all nonempty. For each $x\in K_n$, there is an open neighborhood $U_{x,n}$ which is homeomorphic to an open ball in $\mathbb{R}^m$. Then we have $K_n\subset \bigcup_{x\in K_n} U_{x,n}$. By compactness, there are finitely many $U_{x,n}$ which cover each $K_n$. By unioning these all together for each $n$ gives us a countable covering for $X$ by open sets which are homeomorphic to open balls in $\mathbb{R}^m$. Each of these sets has a countable basis. The countable union of these sets is also a countable set, so it remains to show that this set is a basis, which is easy.