I have tried to prove that a closure of a connected component of the unity in a topological group is closed, but am not sure of its validity. Since it arose from a sentence in a book on the subject,* while the fact that any open subgroup is also closed does not feature there, and since the following depends upon this fact, I must be cautious.
In a (Hausdorff) topological group, denote the connected component of the unity by $\gamma$. Then $\gamma$ is closed.
Denote the closure of $\gamma$ by $\gamma'$, and let $\Omega$ be an open and closed non-empty subset in it. Let $\gamma^0$ be the interior of $\gamma$, and $\Omega'$= $\Omega \cap \gamma^0$. Then $\Omega'$ is closed and open and contained in $\gamma$, and thus $\Omega' = \gamma$ or $\emptyset$, hence $\Omega$ is a closed subset which contains $\gamma$, hence $\gamma' = \Omega$. Therefore $\gamma$ is closed.
An argument, which shows the image of the connected component of the unity in the quotient group is the connected component there, follows the statement; it uses the argument of forming the union of two closed sets without common points. But as we cannot assume in advance that $\gamma$ is open, this argument fails to produce the result here. In the end, I would like to express the thank to those who spend their time reading and answering this question.
*The book is L'intégration dans les groupes topologiques et ses applications by A. Weil.