Closure of a subgroup is again a subgroup Let $G$ be a topological group and $H$ a subgroup. Then $\overline{H}$ is again a subgroup.
Attempt: Let $x,y \in \overline{H}$. Choose nets $\{x_\alpha\}_{\alpha \in I}$ and $\{y_\beta\}_{\beta \in J}$ with $x_\alpha \to x, y_\beta \to y$ and these nets are in $H$. Then we get a net $\{(x_\alpha, y_\beta)\}_{(\alpha, \beta) \in I \times J}$ where $I \times J$ is ordered in the obvious way such that $(x_\alpha, y_\beta) \to (x,y)$ in $G \times G$. By continuity of multiplication, we obtain $x_\alpha y_\beta \to xy$ and since $x_\alpha y_\beta \in H$ for all indices $\alpha, \beta$, we get $xy\in \overline{H}$.
Is this correct?

Seems ok, but I'd use "pre-image of open is open" directly instead of converging nets.
Let $x,y\in\overline H$.
Suppose $xy\notin\overline H$. By openness of $\overline H^\complement$, there is a neighbourhood of $U$ of $(x,y)\in G\times G$ such that $uv\notin \overline H$ for all $(u,v)\in U$. $U$ contains some $V_x\times V_y$ where $V_x,V_y$ are neighbourhoods of $x,y$, respectively. Pick $u\in V_x\cap H$, $v\in V_y\cap H$ and arrive at a contradiction.
