Is every (second-countable) topological manifold completely metrizable?
It is known that every smooth manifold possess a complete Riemannian metric, hence in particular it is completely metrizable, however there are non smoothable manifolds.
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The proof I know is a little roundabout. Let $M$ be your manifold. It is locally compact and Hausdorff, so it has a one-point compactification $M^*$ which is compact Hausdorff. Now $M^*$ is again second countable (see One point compactification is second contable), and (locally) compact Hausdorff spaces are regular, so by the Urysohn metrization theorem, $M^*$ is metrizable with some metric $d^*$. Since $M^*$ is compact, then $(M^*, d^*)$ is of course complete. Now $M$ is an open subset of $M^*$, and every open (or even $G_\delta$) subset of a complete metric space is completely metrizable (with a different metric). See Theorem 1.2 of this note for a proof; it's also in Kechris's Classical Descriptive Set Theory and probably many other standard texts.
In fact, unless I am mistaken, we just showed any locally compact Hausdorff second countable space is completely metrizable.
If there is a more direct proof, I would be interested to see it!