Show that each connected component of a topological space $X$ is connected. 
Show that each connected component of a topological space $X$ is connected.

Here is my proof
My Proof: Choose a connected component $$C = [x] = \{y \in X \mid \exists \text{ a connected set containing both $x$ and $y$}\}$$
Now for each $y \in C$ let $C_y$ be the connected set containing $x$ and $y$. We claim that $C = \bigcup_{y \in [x]}C_y$. To prove this claim pick $\alpha \in \bigcup_{y \in [x]}C_y$, then $\alpha \in C_y$ for some $y \in C$. Since $C_y$ is a connected set in $X$ containing both $\alpha$ and $x$, it follows that $\alpha \in C$ hence $\bigcup_{y \in [x]}C_y \subseteq C$.
Conversely pick $\beta \in C$, then by construction $\beta \in C_{\beta}$ which is the connected set containing both $x$ and $\beta$. Thus $\beta \in \bigcup_{y \in [x]}C_y$ and we have $C = \bigcup_{y \in [x]}C_y$.
Now since $\bigcup_{y \in [x]}C_y$ all contain $x$, their intersection is nonempty and since each $C_y$ is connected it follows that $\bigcup_{y \in [x]}C_y$ is connected. Thus $C$ is connected. $\square$

First of is my proof correct? Can I improve it in any way? I'm not too happy about my construction of $\bigcup_{y \in [x]}C_y$, because I said 

"for each $y \in C$ let $C_y$ be the connected set containing $x$ and $y$"

but there's no need for $C_y$ to be unique and that made matters a bit hairy when I tried to show that $C \subseteq \bigcup_{y \in [x]}C_y$
Is there a better way to prove this?

I don't think it matters too much that the sets $C_y$ may not be unique.  The proof rests on the notion that a union of connected sets with common intersection is connected, which seems plausible (I haven't tried to prove it though).  If that isn't an established proposition in your text though, I think it should be proved.
But the construction seems a little complicated.  You might just appeal to the definition of connected.  If $C = [x]$ were not connected, you could find disjoint sets $X$ and $Y$ such that $C \cap X \neq \varnothing$, $C \cap Y \neq \varnothing$, and $C \subseteq X \cup Y$.  Choose an $x\in X \cap C$ and $y \in Y \cap C$, and a connected set $D$ containing $x$ and $y$.  Can you argue to a contradiction from here?
