Does taking closure preserve finite index subgroups? Let $K \leq H$ be two subgroups of a topological group $G$ and suppose that $K$ has finite index in $H$. Does it follow that $\bar{K}$ has finite index in $\bar{H}$ ?

The answer is yes in general, and here is a proof, which is an adaptation of MartianInvader's:
Let $K$ have finite index in $H$, with coset reps. $h_1,\ldots,h_n$.
Since multiplication by any element of $G$ is a homeomorphism from $G$ to
itself (since $G$ is a topological group), we 
see that each coset $h_i \overline{K}$ is a closed subset of $G$,
and hence so is their union $h_1\overline{K} \cup \cdots \cup h_n \overline{K}$.
Now this closed set contains $H$, and hence contains $\overline{H}$.  Thus
$\overline{H}$ is contained in the union of finitely many $\overline{K}$
cosets, and hence contains $\overline{K}$ with finite index.
[Note: I hadn't paid proper attention to Keenan and Kevin's comments on MartianInvader's answer when I wrote this, and this answer essentially replicates the content of their comments.]
