I'm confused, if you are talking about the fact that a subgroup of a discrete subgroup is discrete, this really is just the (more general statement) that a subspace of a discrete space is discrete. In particular, you want to show that each element of is open, but for each you know (via the fact that is discrete) that there exists some open set such that . Clearly though and so is open in . Since was arbitrary the conclusion follows. Make sense?