Let $G$ be the usual: a topological Abelian group with a topology induced by a countable neighbourhood basis $G_n$ of zero such that $G = G_1 \supset G_2 \supset \dots$. Let $\widehat{G}$ denote the completion of $G$.
Apparently, from $$ \widehat{\widehat{G}} \cong \widehat{G}$$
it follows that if $G$ is complete then it is Hausdorff. Maybe it's just a bit late but it's not obvious to me. So if $G$ is complete we have $G \hookrightarrow \widehat{G}$ so that $G$ is a subgroup of $\widehat{G}$.
We know that if $H$ is the intersections of all neighbourhoods of zero then $G$ is Hausdorff if and only if $H = \{0\}$. Does this help here? Thanks for your help.
$H$ is a subgroup of $G$. If $x\in H$, then $\{nx,n\in\Bbb Z\}\subset H$. In particular, the sequence $\{nx\}_{n\geq 1}$ is Cauchy, and converge to some $y$. Since $\{(n+1)x\}$ converges to $y$, by continuity of the addition $x=0$.