Show that $A\cup C$ and $C\cup B$ are connected 
Question. $\;(X,d)$ is a connected metric space, $C$ is a connected subset of $X$ such that $C^\complement=A\cup B$, where $A$ and $B$ are separated sets. Show that $A\cup C$ and $C\cup B$ are connected.

My approach:
Suppose $C\cup B$ is disconnected. Then there exist opens sets $G_1, G_2$ such that
$$C\cup B\,\subset\, G_1\cup G_2,$$
$$(C\cup B) \cap G_1 \,\not=\,\emptyset\,\not=\,(C\cup B) \cap G_2,$$
$$(C\cup B) \cap G_1 \cap G_2\,=\,\emptyset.$$
Since $C \subset C\cup B$ and C is connected, thus $C\subset G_1$ or $C\subset G_2$.
WLOG, assume $C\subset G_1$ and $C\cap G_2\not=\emptyset.$ Then, $C\cap G_2\subset G_1\cap G_2\not =\emptyset.$
After this point I am unsure of what to do. I looked at the other similar solutions but I do not understand those.
 A: I recommend that you draw a picture for the sets $A,B,C, A',B', U,V, P,Q$ in this answer as much of the set-theoretic calculations then look obvious.
Since $A,B$ are separated, let $A', B'$ be open subsets of $X$ such that $A\subseteq A'$ and $B\subseteq B'$ and $A'\cap B=\emptyset =B'\cap A.$
We show $A\cup C$ is connected. Then by interchanging the letters $A,B$ & also $A',B'$, we also have that $B\cup C$ is also connected.
Suppose   $U,V$ are open subsets of $X$ with $U\cup V\supseteq A\cup C$ and $(U\cap V)\cap (A\cup C))=\emptyset. $ The goal is to show that $$ [G1]....\quad U\cap (A\cup C)=\emptyset$$ or that  $$[G2]....\quad V\cap (A\cup C)=\emptyset.$$
Now $C$ is connected, and we have $C\subseteq U\cup V$ with $U\cap C=\emptyset = V\cap C.$ So $C\subseteq U$  or $C\subseteq V.$ WLOG let $$C\subset V \land  C\cap U=\emptyset.$$
Claim: The two open sets $$P=V\cup B', \quad  Q=U\cap A'$$ are disjoint and that they cover $X$.
Proof of claim: $(i).$ Since $X=A\cup B\cup C,$ we will have $P\cap Q=\emptyset$ if we have $P\cap Q\cap Y=\emptyset$ for each $Y\in \{A,B,C\}.$
We have  $P\cap A=(V\cap A)\cup (B'\cap A)=(V\cap A)\cup (\emptyset)=V\cap A.$ We have $Q\cap A=U\cap A'\cap A=U\cap A.$
So $P\cap Q\cap A=(V\cap A)\cap (U\cap A)\subseteq (U\cap V)\cap (A\cup C)=\emptyset.$
I will  leave the calculations that $P\cap Q\cap B=\emptyset=P\cap Q\cap C$ to the reader.
$(ii).$ To show that $P\cup Q=X=A\cup B\cup C,$ we show that $P\supseteq B\cup C$ and that $P\cup Q\supseteq A.$
Since $V\supseteq C$ and $B'\supseteq B,$ we  have $P=V\cup B'\supseteq (V\cap C)\cup (B'\cap B)=C\cup B .$
Since $A'\supseteq A,$ we have $P\cup Q=(V\cup B')\cup (U\cap A')\supseteq V\cup (U\cap A)\supseteq (V\cap A)\cup (U\cap A)=(V\cup U)\cap A=A$
because $V\cup U\supseteq A\cup C\supseteq A.$
End of proof of Claim.
Now $X$ is connected so $P=\emptyset$ or $Q=\emptyset.$
If $P$ is empty then $V$ is empty because $P\supseteq V,$ and we have $$[G2]'....\quad V\cap (A\cup C)=\emptyset.$$ If $Q$ is empty then since $A\subseteq A'$ and $C\cap U=\emptyset,$ we have $$[G1]'....\quad U\cap (A\cup C)=(U\cap A)\cup (U\cap C)\subseteq (U\cap A')\cup (U\cap C)=Q\cup (U\cap C)=\emptyset.$$
Remarks. Notice this applies to any topological space $X$. This Q (for any $X$) is an exercise in General Topology by R. Engelking, and (as always) he gives the original publication reference for it. There should be a simpler proof for metrizable $X$ because if $X$ is metrizable then we can take $A',B'$ to be disjoint.
