 Solution

Let $x,y\in \overline{H}$ and let $W$ be an open neighbourhood of $x{y}^{-1}$. Continuity of $(u,v)\mapsto u{v}^{-1}$ supplies neighbourhoods $U$ of $x$ and $V$ of $y$ with $U{V}^{-1}\subseteq W$. Choose $h\in H\cap U$ and $k\in H\cap V$. Then $h{k}^{-1}\in H\cap W$, so $x{y}^{-1}\in \overline{H}$. The subgroup criterion using $x{y}^{-1}$ proves the closure of a subgroup is a subgroup. The neighbourhood argument avoids assuming the topology is determined by sequences.