Let $G$ be a finitely generated group.
Recall that a group $L$ is LERF if every finitely generated subgroup $H$ of $G$ is equal to the intersection of the finite index subgroups of $G$ which contain $H$ (i.e. every finitely generated subgroup is profinitely closed).
Assume that $G$ is virtually LERF. That is, there is a finite index subgroup $L$ of $G$ which is LERF. Does this imply that $G$ is LERF? (this is indeed the situation with residually finite groups, but I don't know whether this extends to LERF groups)
Yes, and it's easy.
Let $H$ be a f.g. subgroup of $G$. so $H\cap L$ is intersection of finite index subgroups of $L$ and hence of $G$; that is, $H\cap L$ is closed in the profinite topology. Being a finite union of cosets of $H\cap L$, $H$ is closed as well.