Is every topological manifold completely metrizable? 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.
 A: Yes.  
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!
A: Express $M$ as a union of open sets $\{U_n\}$ with $clos(U_n)$ compact and contained within $U_{n+1}$ (always possible if $M$ is not itself compact).
Put a Riemannian metric $g_0$ on $M$.  Revise $g_0$ to $g_1$ as follows:  $g_1 = f_1 g_0$ for $f_1$ a positive scalar function obeying:
(a) $f_1 = 1$ on $U_1$ and
(b) on $U_3 - U_2$, $f_1 < 1/\sqrt{d_0(U_2, M - U_3)}$ (where $d_0$ = distance function generated by $g_0$).
Note that for $d_1$ the distance function generated by $g_1$, $d_1(U_2, M - U_3)$ is at least $1$; in particular, any curve starting in $U_1$ and escaping to infinity must have at least length $1$, because it crosses from the boundary of $U_2$ to the boundary of $U_3$, a distance of at least $1$.
Now continue inductively: treating $U_3$, $U_4$, and $U_5$ as we just did $U_1$, $U_2$, and $U_3$ and creating $g_2 = f_2 g_1$, with any curve from $U_1$ escaping to infinity having length at least $2$, as it crosses both from boundary of $U_2$ to $U_3$ (same metric as $g_1$) as well as from boundary of $U_4$ to boundary of $U_5$ (again with distance at least $1$); and so on.
We end with a well-defined metric $g_{\infty} = \lim g_n$.  Any curve escaping to infinity must cross an infinite number of bands of width at least $1$ in $g_{\infty}$, so it must have infinite length in $g_{\infty}$.  That means $(M, g_{\infty})$ is complete.
