Every maximal proper closed subgroup of a profinite group is open

Choose $x\in G\setminus H$. By [L1], write $G$ as an inverse limit of finite groups and view $H$ as a closed subset of that product. Since $x\mathrm{\notin }H$, some cylinder neighbourhood of $x$ misses $H$; equivalently, there is an open normal subgroup $N$ of $G$ such that $xN\cap H=\mathrm{\varnothing }$.

The subgroup $HN$ is open because it is a union of $N$-cosets, and it is closed because it is a finite union of closed cosets. If $HN=G$, then $x=hn$ for some $h\in H$ and $n\in N$, so $h=x{n}^{-1}\in xN\cap H$, contradicting step 1.1. Thus $HN$ is a proper closed subgroup containing $H$. By maximality of $H$, one must have $HN=H$, hence $N\le H$. Therefore $H$ contains an open neighbourhood of the identity and is itself open.