Theorem - Every subgroup of a topological group is either clopen or has empty interior.

$\,$

Proof: Let $G$ be a topological group and $H\subseteq G$ a subgroup. Suppose the interior of $H$ is nonempty, i.e. there is a non-empty open set $U$ of $G$ such that $U\subseteq H$. Translating $U$ around $H$ we can see that $H$ is open: if $u\in U$ then for every $h\in H$ the set $hu^{{-1}}U$ is open in $G$, is contained in $H$ and contains $h$, which implies that $H$ is open in $G$.

Let us now see that $H$ is closed. Let $\overline{H}$ denote the closure of $H$ and let $H^{2}$ be the set of elements of the form $h_{1}h_{2}$ where $h_{1},h_{2}\in H$. Of course, since $H$ is a subgroup of $G$, we have that $H^{2}=H$. Also, since $H$ is open we know that$H\subseteq\overline{H}\subseteq H^{2}$ (see this entry - Proposition 5). Hence $\overline{H}=H$, i.e. $H$ is closed.

We have proven that a subgroup of a topological group must be clopen or it must have empty interior. Since this two topological properties can never be satisfied simultaneously, we have that every subgroup of a topological group is either clopen or it has empty interior. $\square$