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.

Isn't the complement of $A \cup B$ just C? And I think it's easy to see that C is open.
