Proof:

Let $U$ be the open neighborhood of the identity that is contained in $H$ and $h \in H$ be arbitrary. Since multiplication by $h$ is a homeomorphism, the set $Uh = \{uh / u\in U\}$ is also open and also contained in $H$. Hence the set

$$L = \bigcup_{h\in H} Uh$$

is open and contained in $H$. Since $U$ contains the identity element, $H \subset L$ and we conclude that $H$ is open. Our job will be over if we can show that $H$ is closed as well. Let $x \in \overline{H}$ be arbitrary. Since the neighborhood $Ux$ of $x$ contains a point $y \in H$, there exists $u \in U$ such that $y = ux$ which, in view of the fact that $U \subset H$, implies $x \in H$. Hence $\overline{H} = H$. $\square$