Let $G$ be a topological group and let $H$ be an open subgroup of $G.$
Assume by contradiction that $H$ is not closed. Then, there exists a convergent net $(h_\alpha)_{\alpha \in I}$ of points of $H$ such that $$h_\alpha \longrightarrow \bar h$$
and $\bar{h} \notin H$. So $\bar{h} \in G \setminus H$, which is closed. Thus, there exists a convergent net $(h^c_\alpha)_{\alpha \in I}$ of points of $G \setminus H$ such that $$h^c_\alpha \longrightarrow \bar{h}.$$
It follows that $$h_\alpha(h^c_\alpha)^{-1} \longrightarrow \bar{h}\bar{h}^{-1} = 1_G.$$
But $H$ is a open neighborhood of $1_G$ which contains no element of the form $hh^c$, where $h\in H$ and $h^c \in G \setminus H$. Contradiction. Thus, $H$ can only be closed.
Is the reasoning correct or is there something I can improve on? I'm studying nets and have fun proving some statements using them.
It is correct but can be simplified by taking $h^c_α=\bar h.$ The fact that $G\setminus H$ is closed was anyway not used to construct your net $(h^c_α).$ I.e. the word "Thus" between "is closed" and "there exists" was undue.
Another improvement (imo) is to avoid reasoning by contradiction.
Let $H$ be an open subgroup of $G,$ and $(h_\alpha)_{\alpha\in I}$ a net in $H,$ converging to some $h\in G.$ Let us prove that $h\in H.$
$k_\alpha:=h_\alpha h^{-1}\to1\in H.$ Since $H$ is open, we deduce $k_\alpha\in H$ for some $\alpha,$ whence $h=k_\alpha^{-1}h_\alpha\in H,$ q.e.d.