TITLE: Connected Components are Closed
QUESTION [5 upvotes]: I am trying to prove that if $C$ is a connected component of $X$, then $C$ is closed. Here is my attempt:
Let $C$ be a connected component of $X$. Then $\overline{C} \supseteq C$ must be connected as well, and since every connected subspace intersects one, and only one, connected component (and is therefore contained in it), it follows that $\overline{C} \subseteq C$. 
Is this right? Something about it is fishy...

REPLY [7 votes]: Your proof is about right.
I'd formulate it as follows: a connected component $C$ of $X$ is a maximally connected subset; this means 2 things:

$C$ is connected.
if $C \subseteq D$ and $D$ is connected, $C=D$.

Now use that $C$ connected implies $\overline{C}$ connected.
Then applying 2. and noting that obviously $C \subseteq \overline{C}$, we conclude that $C = \overline{C}$, which is equivalent to $C$ being closed. QED.