Connected set and connected subset question Define $A,B$ to be separated sets if $A \cap \overline{B}=\overline{A} \cap B=\varnothing$.
Let $X$ be a connected space, $C$ a connected subset of $X$ and $X-C=A \cup B$ where $A,B$ are separated sets. Prove $A \cup C$ and $B \cup C$ are connected.
Here is my effort:Suppose one of $A \cup C$ or $B \cup C$ is not connected. Assume $A \cup C$. Let $U,V$ be a pair of disjoint, nonempty open sets with union $A \cup C$. Then $C \subset A \cup C \subset U \cup V$. So that $C \subset U$ or $C \subset V$. Assume $C \subset V$. Then $X-V \subset X-C=A \cup B$. At this point I do not think I am going in the right direction. I can't figure out where to go next with this. How can I take the right approach to solve this problem?
I saw a proof for this that was somewhat technical and would like to know if there is any way my approach could end up with a correct proof.
 A: Here is one way to start as you did and finish the proof :

Suppose one of $A \cup C$ or $B \cup C$ is not connected. Assume $A \cup C$. Let $U,V$ be a pair of disjoint, nonempty open sets with $A \cup C\subset U\cup V$.

Since $A\cup C$ is not assumed to be open in $X$, this is the best we can do.

Then $C \subset A \cup C \subset U \cup V$. So that $C \subset U$ or $C \subset V$. Assume $C \subset V$.

Then let $U' = U - \overline{B}$ and $V' = V \cup (X - \overline{A})$. Both are open sets.
Since $A$ and $B$ are separated, we have $A \subset U'$, $B\subset V'$ and $C\subset V \subset V'$. So $X = U'\cup V'$.
Furthermore, $U' \cap V' = (U- \overline{B}) \cap (X- \overline{A}) \subset U \cap C \subset U\cap V = \emptyset$.
Since $X$ is connected, this is absurd (I am assuming here that $A,B,C$ are non-empty. There is a little more work to deal with the special cases)
Another approach using a characterization of connected spaces which is often useful  :
Let $f: A\cup C \to \{0,1\}$ be a continuous map (where $\{0,1\}$ has the discrete topology).
Then, $f|_C$ is constant (wlog, let's take it equal to $0$). Then, since $A$ and $B$ are separated, we can extend $f$ to a continous map $g:X\to \{ 0,1\}$ by setting $f|_B = 0$. Since $X$ is connected, $g$ must be constant and also $f$. Therefore, $A\cup C$ is connected
