Proposition 4

Let $G$ be a topological group and let $H$ be a subgroup of $G$ which is locally closed at one point of $H$ (Chapter I, § 3, no. 3, Definition 2). Then $H$ is closed in $G$.

By translation, $H$ is locally closed at each of its points, i.e. $H$ is locally closed in $G$. Let $V$ be a symmetric open neighbourhood of $e$ in $G$ such that $V \cap H$ is closed in $V$. If $x \in \overline{H}$, then $xV$ meets $H$; if $y \in xV \cap H$, we have $x \in yV$, and $y(V \cap H) = (yV) \cap H$ is closed in $yV$. But $x$ lies in the closure of $(yV) \cap H$, hence $x \in H$.