I've found some literature which would be helpful if I understood the following,
"we can choose neighborhood $U,V$ of the identity such that $VV^{-1} \subset U$ and $U \cap \Gamma = \{e\}$. "
Question: I can figure out the last, but how do you choose set such that $VV^{-1} \subset U$?
So that this question does not go unanswered: picking $V = U\cap U^{-1}$ does the trick. This is a standard trick used in the theory of topological vector spaces as well as in harmonic analysis when proving very technical lemmas.
Edit: The above is incorrect as I had jumped the gun a bit. Taking the intersection of a set with its inverse is necessary to the proof, but there is more to it since we have to whittle down the set a bit to keep from leaving it when we multiply by the inverse set.
Since multiplication is continuous and $U$ contains $1$, consider the inverse image $\mu^{-1}(U)$, where $\mu$ is the multiplication map. Since $U$ is open, $\mu^{-1}(U)$ is open. Moreover, $(e,e)\in \mu^{-1}(U)$. Let $V_1,V_2$ be open neighborhoods of $e$ such that $V_1V_2\subseteq U$. Taking $V_3 = V_1\cap V_2$ and $V = V_3\cap V_3^{-1}$, we have the result.