closed subgroup on a topological group

First, a bit of point set topology: $\bar U \cap H$ closed is equivalent to $U \cap H$ is closed in $U$.

Second, by homogeneity, for every $h\in H$, there exists $U_h$ neighborhood of $h$ so that $U_h \cap H$ is closed in $U_h$ ( take $U_h = h\cdot U$)

The above means that $H$ is locally closed in $G$. This is equivalent to : $H$ is the intersection between a closed set and an open set, or, what we are really after: $H$ is open in its closure $\bar H$.

Now, $\bar H$ is a topological group ( with the induced topology. Let's now recall: an open subgroup of a topological group is also closed. Indeed, any coset will be open, so any union of cosets. Therefore $H$ is also closed in $\bar H$, and so equal to $\bar H$, closed.

Take home message: a locally closed subgroup is closed.

Suppose $x\in\overline{H}$. Let $V$ be a neighborhood of $1$ such that $VV^{-1}\subseteq U$. Since $Vx$ is an open neighborhood of $X$, $x\in\overline{H\cap Vx}$. Now pick some $y\in H\cap Vx$ and multiply everything by $y^{-1}$ on the right. We conclude that $xy^{-1}\in\overline{Hy^{-1}\cap Vxy^{-1}}$. Since $H$ is a subgroup and $y\in H$, $Hy^{-1}=H$. Also, since $y\in Vx$, $xy^{-1}\in V^{-1}$ so $Vxy^{-1}\subseteq VV^{-1}\subseteq U$. Thus $$xy^{-1}\in\overline{H\cap U}\subseteq \overline{H\cap\overline{U}}=H\cap \overline{U}.$$ In particular, $xy^{-1}\in H$, so since $y\in H$ and $H$ is a subgroup, $x\in H$.