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 $\bar{H}$ ?

Yes, if your group has a countable basis (so the closure of a set is the set of limits of sequences of points in that set).  Let $\{h_1,...,h_n\}$ be representatives of the left cosets of $K$ in $H$.  Then for any point $h \in \bar{H}$, express it as a limit of a sequence $x_1, x_2, x_3, ...$ of points $x_i \in H$.  These points can then be represented using the coset representatives as $h_{i_1}y_1, h_{i_2}y_2, h_{i_e}y_3,...$, with $i_k \in \{1,...,n\}$ and $y_n \in K$.
Now some $h_k$ must appear infinitely often, thus by passing to a subsequence we have that $h$ is the limit of a sequence $h_ky_1, h_ky_2, h_ky_3,...$ and so by the continuity of group multiplication we have $h = h_ky$, where $y \in \bar{K}$ is the limit of the $y_i$.
This shows that any point $h \in \bar{H}$ lies in $h_k\bar{K}$ for some $k \in \{1,...,n\}$, thus the same coset representatives of $K$ in $H$ also form a set of (possibly redundant) coset representatives of $\bar{K}$ in $\bar{H}$, and in particular $\bar{K}$ has finite index in $\bar{H}$.
I suspect the sequences in this argument could be adapted into more of a "for any open set around $h$, there is ..." sort of language to make the proof work for general topological groups, but I prefer the intuition you get with sequences.
