A group is residually finite iff profinite topology is Hausdorff

Question

In solving some exercises in Sageev's notes on $CAT(0)$ cube complexes I've needed to use the following fact that I would like to know how to prove.

A group $G$ is residually finite if and only if the profinite topology on $G$ is Hausdorff.

Here the working definition of residually finite is that the trivial subgroup is separable; where a subgroup $H < G$ is separable if for all $g \in G-H$ there is a finite index $K < G$ with $H<K$ and $g \not\in K$.

The profinite topology is generated by the basis of cosets of finite index subgroups of $G$. It can also be shown that you could instead use cosets of finite index normal subgroups of $G$.

Any hints are appreciated, and I actually only need to understand the forward direction.

Answer 1

Hint: if $x,y \in G$ with $x \neq y$, then $xy^{-1} \neq 1$, hence there is an open normal subgroup $N$ (of finite index) with $xy^ {-1} \notin N$. But then $xN \cap yN = \emptyset$.