If A complement is the union of two separated sets, prove that the union of those separated sets with A is connected. Let $A$ be a connected subset of a connected metric space $(X,d)$.
Assume $A^{c}$ is the union of two separated sets $B$ and $C$.
Prove that $A \cup B$ and $A \cup C$ are connected.
Attempt
Proving $A \cup B$ is connected is sufficient.
Assume towards a contradiction that $A \cup B$ is not connected. (So it is disconnected).
There exists open sets $G_{1}$ and $G_{2}$ such that $A \cup B \subseteq G_{1} \cup G_{2}$, $(A\cup B) \cap G_{1} \neq \phi$, $(A\cup B) \cap G_{2} \neq \phi$ and $(A\cup B)\cap G_{1} \cap G_{2} = \phi$.
Since $A \subseteq A \cup B$ and $A$ is connected, then either $A$ lies in $G_{1}$ or $G_{2}$.
WLOG, assume $A$ lies in $G_{1}$.
I do not know how to proceed from here. I need to obtain a contradiction to finish my proof.
Edit
I have corrected my proof.
 A: You are almost there. If $A$ lies entirely in $G_1$, and $A\cap G_2\neq \emptyset$, then you can easily conclude that $G_1\cap G_2$ cannot be an empty set, therefore, a contradiction follows.
You can conclude this because you know that $A\cap G_2\subseteq G_1\cap G_2$, and $\emptyset$ has no non-empty subsets.
A: Observation:if $X$ is disconnected, then there exists non-empty subsets $U, V$ which, in addition to satisfying that $X=U \cup V$, satisfy the following equivalent conditions:


*

*$U, V$ are disjoint and $U, V$ are open in $X$

*$U, V$ are disjoint and $U, V$ are closed in $X$

*$U$ $\cup $ $cl(V)_X = \emptyset$ and $V$ $\cup $ $cl(U)_X = \emptyset$
Lemma:suppose $X$ is disconnected, so that there exists non-empty subsets $U, V$ such that $X=U \cup V$, $U, V$ are disjoint, and $U, V$ are open in $X$. If $A$ is a connected subset of $X$, then $A\subseteq U$ or $A\subseteq V$. 
Suppose, toward contradiction, that $A\cup B$ is disconnected, so that there exists non-empty subsets $G, H$ such that $A\cup B=G \cup H$, $G, H$ are disjoint, and $G, H$ are open in $A \cup B$.
Since $A$ is connected, w.l.o.g. suppose that $A\subseteq G$, so that $H \subseteq B$. 
Claim 1: $H$ is closed in $X$.
Proof 1: 
Since $B$ is closed in $X\setminus A$ and $cl(B)_{B\cup C}\cap C=\emptyset $, $cl(H)_{A\cup B\cup C}=cl(H)_{A}\cup cl(H)_{B\cup C}=cl(H)_{A}\cup cl(H)_{B}=cl(H)_{A}\cup H$.
If $x\in cl(H)_{A}\setminus H$, then $x\notin H, x\in A$, so that $x\in G$. This implies $x \in cl(H)_A\cap G \subseteq cl(H)_{A\cup B}\cap G$, so that $cl(H)_{A\cup B}\cap G \neq \emptyset$, which is a contradiction. Therefore, $cl(H)_A=H$. 
Putting everything together, we see that $cl(H)_{A\cup B\cup C}=cl(H)_{A}\cup H=H \cup H=H$. Therefore, $H$ is closed in $A\cup B\cup C=X$.
Claim 2: $H$ is open in $X$.
Proof 2: 
If $x\in H$ and $x\notin int(H)_{X}$, then every neighbourhood of $x$ contains some point in $X\setminus H=(A\cup B \cup C)\setminus H$, so that every neighbourhood of $x$ contains some point in $C\setminus H$. This implies that $x\in cl(C)_{B\cup C}$, so that, since $x\in H\cap cl(C)_{B\cup C}\subseteq B\cap cl(C)_{B\cup C}$, $B\cap cl(C)_{B\cup C}=B\cap cl(C)_{X\setminus A}\neq \emptyset$, which is a contradiction. Therefore, $H=int(H)_{X}$, so that $H$ is open in $X$. 
Conclusion: since $H$ is a nontrivial clopen subset of $X$, $X$ is disconnected, which is a contradiction. Therefore, the supposition that $A\cup B$ is disconnected must be false. 
A: Isn't the complement of $A \cup B$ just C? And I think it's easy to see that C is open.
