I'm trying to understand a proof of a theorem but I didn't understand a point.
Let $G$ be an locally compact abelian group. Denote $G_0$ the connected component of $0$ (the identity of $G$). It's an open subgroup of $G$. Indeed for every $x\in G_0$, $x-G_0$ is connected and intersects $G_0$, so $x-G_0\subset G_0$. (In other words for every $x,y\in G_0$, $x-y\in G_0$.) The quotient group $G\setminus G_0$ is an locally compact abelian totally disconnected group. We can show under these conditions that $G\setminus G_0$ admits an open compact subgroup $K$. Let $\phi $ the canonical projection from $G$ to $G\setminus G_0$ and put $G_1=\phi ^{-1}(K)$. $G_1$ is an open subgroup of $G$. Why every open subgroup of $G_1$ must contain $G_0$ ?
Assume that $H\subset G_0$ is an open subgroup of $G$ contained in the connected component of the identity $G_0$.
Then write $G_0$ as a disjoint union of cosets $G_0 = \bigsqcup_{g} (g+H)$, for some set of coset representatives. Since $H$ is open, so is each coset $g+H$.
Now if $H\subsetneq G_0$, then
$G_0$ = $H \sqcup \bigsqcup_{g\neq 0} (g+H)$ writes $G_0$ as a disjoint union of open subsets, contradicting connectedness.
More generally, every open subgroup $H\subset G$ is also closed, so disconnects $G = H \sqcup \bigsqcup_{g\neq 0} (gH)$, where $G$ is any topological group.