A subset $C$ of a pro-finite group is closed iff $C$ is compact

Question

I'm trying to prove both directions..

I think I managed to prove one: if $C$ is compact then $C$ is closed.
Feel free to correct me:

1. Every pro-finite group is compact, Hausdorff, and has a basis for its topology consisting of open-closed sets.
2. $C$ is compact then any open cover of $C$ contains a finite open sub-cover.

Derived from both statements: $C=\bigcup_{i\in I}B_i$ (finite union) where $B_i$ is a basis element of the group.
Because each basis element is open-closed, and the union is of finite elements, $C$ is closed as it's a finite union of closed sets.

I'd appreciate assistance for the other direction.

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

As s.harp said in his comment, profinite groups are compact Hausdorff. In compact Hausdorff spaces closed subspaces agree with compact subspaces.