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

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.

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

-
I don't think your argument is correct as written. For example, I can take $\Omega=\emptyset$, which is certainly a clopen subset of $\gamma'$, but it need not be true that $\Omega'=\gamma$. –  Arturo Magidin Nov 27 '11 at 4:13
@ArturoMagidin: Yes, you are right. In the argument, perhaps it should be said that $\Omega'$ is either $\emptyset$ or $\gamma$. But this might not affect the argument much, can it? Thanks for your pointing this error out. –  awllower Nov 27 '11 at 6:06
If the intersection is empty, then it cannot be the case that $\Omega$ contains $\gamma$ unless the interior of $\gamma$ is empty; even if the interior is empty, you have no warrant for asserting that $\Omega$ contains $\gamma$ as it currently stands. –  Arturo Magidin Nov 27 '11 at 6:12
@ArturoMagidin: I see the point now. Then how can one show this proposition? Or I have per chance to try some degree harder? Could you help me? Thank you very much. –  awllower Nov 27 '11 at 6:37
@ArturoMagidin: Wait, since $\gamma'$ is the closure of $\gamma$, that intersection cannot be empty, can it? As it must contain some points in the closure, it absolutely has to have a non-empty intersection with $\gamma$ by definition. –  awllower Nov 27 '11 at 6:39

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.
Eh, basically I am trying to work out the details here. And the argument above in fact is trying to prove that the closure of the connected component is connected as well. In addition, how to show that $Z$ is closed? Thanks in any case for the answer. –  awllower Nov 27 '11 at 6:43
@awllower Yes. For a subset of $X$ to be connected we mean that it is connected in the subspace topology. I was only trying to emphasize that while $A$ is closed in both $X$ and $\overline Y$, $A \cap Y$ might only be closed as a subset of $Y$. –  Dylan Moreland Nov 27 '11 at 16:15