$\sigma$-compactness implies second countability

Question

I’m working through Introduction to Smooth Manifolds by John M. Lee. It begins with a few topology exercises, and I need help knowing where to start on one of them. The question is:

Prove that a locally Euclidean Hausdorff space is a topological manifold iff it is $\sigma$-compact.

This can be reworded as showing a locally Euclidean, Hausdorff space is second-countable iff it is $\sigma$-compact. I was able to prove that second countability implies $\sigma$-compactness. However, I am unsure how to prove the other direction. I know that because the space is $\sigma$-compact, it must have a countable cover of closed (from the Hausdorff Property) and compact sets. I imagine the proof probably involves using the locally Euclidean property and constructing a basis from that, but I am unsure.

Is this along the right track? How would I proceed from this point? To be clear, I don’t want the proof written for me, rather, I need help knowing how to proceed / a sketch of the proof.

If a space is $\sigma$-compact then every open cover has a countable subcover. If it's also locally Euclidean then it's "locally second-countable", so it's covered by (countably many) second-countable open subspaces. — bof, Apr 07 '23 at 02:57
So it essentially comes down to showing that $\sigma$-compactness implies countable many locally Euclidean neighborhoods? — user_not_found, Apr 07 '23 at 02:59
Does this answer your question? Locally Euclidean Hausdorff topological space is topological manifold iff $\sigma$-compact. — Anne Bauval, Apr 10 '23 at 07:27

$\sigma$-compactness implies second countability

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