Proof:

Let $G_0$ be the connected component of $G$ containing the identity and $h, k \in G_0$ be arbitrary. The set $h^{-1}G_0$ is connected and contains the identity and so $G_0\cup h^{-1}G_0$ is also connected. Since $G_0$ is a component, we have $G_0\cup h^{-1}G_0 = G_0$ which implies $h^{-1}G_0 \subset G_0$. In particular $h^{-1}k$ belongs to $G_0$ from which we conclude that $G_0$ is a subgroup.

Interesting properties of topological groups arise in connection with quotients: