Closure of the connected component of the unity is connected: is my proof valid?

Question

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.

Statement:
In a (Hausdorff) topological group, denote the connected component of the unity by $\gamma$. Then $\gamma$ is closed.

Try:
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.
C.Q.F.D.

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.

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*The book is L'intégration dans les groupes topologiques et ses applications by A. Weil.

Answer 1

Maybe the extra hypotheses are clouding things: in any topological space, connected components are closed. This follows from the fact that if $Y$ is a connected subspace of $X$ then $\overline Y$ is also connected.

Here's (most of) a proof of this last fact. If $Y$ is empty then there is nothing to do, so assume otherwise. Now, if $\overline Y$ is the disjoint union of two closed subsets $A$ and $B$ then one of these, let's say $A$, must intersect $Y$. As $Y$ is the disjoint union of relatively closed subsets $A \cap Y$ and $B \cap Y$, we must have $A \cap Y = Y$ because $Y$ is connected. Does this look promising? Let me know if I should say more.