The normal definition of subgroup separability is:
A group $G$ is said to be subgroup separable if for every finitely generated subgroup $H\leq G$ and $g\in G\setminus H$ there exists a subgroup of finite index $K_g\leq G$ such that $H\leq K_g$ and $g\notin K_g$.
Why is this equivalent to the following?
$G$ is subgroup separable if every finitely generated subgroup of $G$ is closed in the profinite topology of $G$.
The profinte topology is the topology generated by normal subgroups of finite index and there cosets as open sets.
It is easy to show, that the first definition implies the second:
$$H=\bigcap_{g\notin H}K_g$$ We can assume that $K_g$ is normal for all $g\notin H$. So these are open normal subgroups of finite index. Hence they are closed. So the intersection is an intersection of closed sets in the profinite topology and hence closed. So it follows that $H$ is closed.
But how does the second implies the first?
Thanks for help.