# Help understanding proof of every topological group is regular

I found a proof that any topological group is regular here, but I got lost in the last part. The whole argument goes like this:

Consider the map $$f:G \times G \to G$$ defined by $$f(a,b)=ab^{-1}$$. This map is always continuous in a topological group. Now take $$x \in U$$, where $$U$$ is an open set in $$G$$. Then $$f^{-1}(U)$$ contains $$(x,e)$$, so we have $$(x,e)\in V \times W \subseteq f^{-1}(U)$$ for some open subsets $$V$$ and $$W$$ such that $$x\in V$$ and $$e\in W$$.

Hence, $$x\in V$$. Furthermore, $$V\cap (X-U)W=\emptyset$$, since any element in the intersection corresponds to $$a\in V$$, $$b\in W$$, such that $$ab^{-1}\notin U$$, which is a contradiction.

Since $$(X-U)W$$ is an open set containing $$X-U$$, $$X$$ is regular.

The boldface argument is what I don't understand. How can I see that $$V\cap (X-U)W=\emptyset$$?

If $y\in V\cap (X - U)W$, then $y\in V$ and $y = st$ for some $s\notin U$ and $t\in W$. Then $(y,t)\in V\times W$, so $f(y,t) \in U$, i.e., $yt^{-1}\in U$. On the other hand, $s = yt^{-1}\in U$. This contradiction shows $V\cap (X - U)W = \emptyset$.