Symmetric open neighborhood in topological group I want to prove that

($\ast$) If $G$ is a Hausdorff topological group, then for every neighbourhood $U$ of $e$, there exists a symmetric neighbourhood $V$ of $e$ ($V^{-1}=V$), s.t $V*V \subset U$

I had seen a lot related proofs regarding preimage of the map $(x,y)\longmapsto x^{-1}y$ for $G \times G \longrightarrow G$ or $U\times U \longrightarrow G$ but I can't understand them.
For example:


*

*http://www.math.wm.edu/~vinroot/PadicGroups/topgroups.pdf (Prop 1.1 on p2)

*In topological groups. Is every neighborhood of $e$ supset of a square of a symmetric neighborhood of $e$?

*Is a discrete subgroup of a Hausdorff group closed?
My problem is they all take the premimage $\overline U$ of $U$ for $(x,y)\longmapsto x^{-1}y$, but how then the open set $W$ is obtained so that $xy^{-1}\in{U}\forall x,y\in W$? 
I mean $\overline U$ is of the form $\bigcup (x,y)\subset G\times G$, and taking projections $\pi_1(\overline U),\pi_2(\overline U)$ doesn't work, so I am stuck.
So how to proof ($\ast$)? My goal is to prove it, hence every other methods without using preimage are also welcome.

Let $\mu$ denote the multiplication. By continuity, $\mu^{-1}(U)$ is a neighbourhood of $(e,e)$, that means there are [open, if desired] neighbourhoods $W_1,W_2$ of $e$ with $W_1\times W_2 \subset \mu^{-1}(U)$. Now let $W := W_1 \cap W_2$. As an intersection of two neighbourhoods of $e$, it is again a neighbourhood of $e$. Next, let $V := W \cap W^{-1}$. Then $V$ is a symmetric [open, if $W_1,W_2$ are open] neighbourhood of $e$ with $\mu(V\times V) \subset \mu(W_1\times W_2) \subset U$.
