I was going through the open mapping theorem (for topological groups) when I stumbled upon a topological property that I couldn't prove to myself. If I have a topological group, $G$ which is $\sigma$-compact and locally compact, then for every $U=U^{-1}$, a symmetric neighborhood of $e$, there is a precompact neighborhood $V=V^{-1}$ of $e$. Moreover, $V$ can be chosen so that if $W=\overline{V}$ is the closure of $V$, then
$$W^{-1}W\subseteq U.$$
I know I can always find a $V$ as mentioned, but I don't know the reason why its closure should be compact or why the above multiplication should work for the closures. Any help to shed light will be much appreciated.
You do more or less the same thing as proving the initial $V$ exists, it's just a slight modification. Select a pre-compact neighborhood $V'$ of $e$. Then consider $V''=V'\cap U$. Symmetrizing $V''$ by examining $V_0=V''\cap (V'')^{-1}$, we can see that $V_0$ is pre-compact and symmetric since it is contained in $W'$.
Now for containment in $U$ we do the same trick as we do for the original $V$, we find another neighborhood $V_1$ such that $V_1^2\subseteq U$ and set
$$V=V_0\cap (V_1\cap V_1^{-1}).$$
This is still obviously symmetric (that's why I put the parentheses in the intersection) and pre-compact--the closure is a closed subset of $\overline{V'}$ which is compact--and such that $VV^{-1}\subseteq U$--since
$$VV^{-1}=V^2\subseteq V_1^2\subseteq U.$$