Complement of a Component is Connected Given: $X$ is locally connected, $C \subseteq X$ is closed and connected, $U$ is a component of $X \backslash C$.
I want to show that $X \backslash U$ is connected. Here's what I know:
$C$ is closed $\implies X \backslash C$ is open $\implies U$ is open since $X$ is locally connected $\implies X \backslash U$ is closed.
If I suppose that $X \backslash U$ is disconnected, then there are disjoint closed sets $V,W$ so that $X \backslash U \subseteq V \cup W$. 
Since $U$ is merely one component of $X \backslash C$, we have $X \backslash C = U \cup \bigcup U_\alpha$ where each $U_\alpha$ is a component of $X \backslash C$. Then $X \backslash U = C \cup \bigcup U_\alpha$.
I would just like some direction on where to go from here to arrive at a contradiction.
 A: This is false. Consider
$$X=\{0\}\cup\{1\}\cup\{2\}\qquad C=\{0\}\qquad U=\{1\}\qquad X\setminus U=\{0\}\cup\{2\}$$
with the discrete topology.
In view of Stefan H.'s comment, this is still false if $C$ is not open and $U$ is not a component of $X$. Take
$$
X=[0,2]\cup\{3\}\qquad C=[0,1]\qquad U=(1,2]\qquad X\setminus U=[0,1]\cup\{3\}
$$
with the topology induced by the standard topology on $\mathbb{R}$.
A: It would be right if $X$ were connected. So let us assume $X$ is connected.
So $C$ is closed and connected, and $U$ is a component of $X-C$. Since $X$ is locally connected, $U$ is open being a component of an open set in a locally connected space. $U$ is also closed in $X-C$, so there is a closed set $D$ such that $D-C=U$. Without loss of generality we can take $D=U\cup C$. If $X-U=A\sqcup B$ where both $A$ and $B$ are closed, then $C\subseteq A$ since $C$ is connected. Therefore $A\cup D=A\cup U=X-B$ is closed. Thus $B$ is closed and open and must be empty. This implies that $X-U$ is connected.
Edit: I found this post where they show that one can drop the condition that $C$ is closed and $X$ is locally connected. You merely need that $X$ and $C$ are connected and $U$ is a component of $X-C$.
