This question is very simple, but I don't get the right idea.
Assume $H$ be a locally closed dense subgroup of a topological group $G$. Prove that $H=G$.
I need to prove that $gH\cap H\ne\emptyset$ for all $g\in G$. I know that $H$ is open in $\overline{H}$, but I don't know of it is important.
Actually you don't need the two assumptions on $H$, namely that $H$ is locally closed AND that it is open, because each one of these assumptions, separately, imply that $H$ is closed. Proofs of both these assertions can be found here. Now since $H$ is also dense in $G$, its closure coincides with $G$, hence $H=G$.