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?