If $Y$ and $Y \cup X$ are connected. Must their be some $X$-component ($C$) where $C \cup Y$ is connected? This is a question I had, while trying to solve a homework problem. My original approach was dependent upon the following statement being true.

If $Y$ and $Y \cup X$ are connected, then there is some connected component of $X$, call it $C$, where $Y \cup C$ is connected. 

I eventually solved the homework problem using a very different method, but the question has been bugging me. I can't seem to come up with a proof or counterexample.

I haven't been able to make more than trivial progress (for example, if $Y$ is not closed or there is some component $C$ is not closed. Then $\overline{C} \cap \overline{Y} \neq \emptyset$ and you can show that one must contain a limit point of the other, therefore $Y \cup C$ is connected.)

Does anyone have any insights?
The Knaster-Kuratowski fan is a counterexample: let $Y$ be the apex and $X$ the rest of the space. Now, $Y$ is a point and thus connected, and so is $Y\cup X$. But components of $X$ are individual points. So, $Y\cup C$ is disconnected for every component $C$ of $X$, as it is a two point Hausdorff space.

(I imagine there should also be simpler counterexamples. If I think of one, I'll let you know.)
There certainly is some connected component of $X$ where $Y \cup C$ is connected. If there wasn't, then you would get a disjoint union of $X$ and $Y$, a contradiction..

Proof. Choose an arbitrary $x$ in $C$ with the restriction it is the closest point to $Y$ (for some point $y \in Y$). For it to be connected there would have to exist an $\epsilon &gt; 0$ such that $d(x, y) &lt; \epsilon$. If there were no such component, that is, if $d(x, y) &gt; \epsilon$ for all $x$ in $C$ (and since $C$ is an arbitrary subset of $X$, we can just claim $x \in X$ WLOG) then $X$ and $Y$ are clearly disconnected.