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?

This is just an expansion of mathcounterexamples.net's answer above.
Proof: Choose a connected component $C=[x]$. Suppose that $C$ is disconnected, then $C = C_1 \cup C_2$ where $C_1$ and $C_2$ are open in $C$ and disjoint and both are nonempty. Without loss of generality assume $x \in C_1$
We claim that $C_2 = \emptyset$. To prove this claim suppose that $C_2$ was non-empty, then pick $y \in C_2$, then by definition of a connected component (since $y, x \in C$ implies $y \sim x$) there exists a connected set $C_y$ containing both $y$ and $x$. 
Note that $C_y \subseteq C$, to see why this is true, pick $\alpha \in C_y$, then by definition of the relation $\sim$, it follows that $\alpha \sim y$ (since $C_y$ is a connected set in $X$ containing both $\alpha$ and $y$) and since $y \sim x$ and the fact that equivalence relations are transitive it follows that $\alpha \sim x$ which implies that $\alpha \in [x] = C$. Hence $C_y \subseteq C$.
Now since $C_y$ contains both $x \in C_1$ and $y \in C_2$ and $C_y$ is connected and contained in $C$, it follows that $C_1 \cap C_y$  and $C_2 \cap C_y$ is open in $C_y$ and furthermore $C_1 \cap C_y \neq \emptyset$ (since both sets contain $x$) and $C_2 \cap C_y \neq \emptyset$ (since both sets contain $y$) and furthermore $(C_1 \cap C_y) \cup (C_2 \cap C_y) = C_y$ (since $C_1 \cup C_2 = C$ and $C_y \subseteq C$), thus $C_y$ is disconnected a contradiction. Thus $C$ must be connected. $\square$ 
