Prove that C is connected whenever there exists a connected subset.

Question

Let $(X,d)$ be a metric space and $C \subseteq X$ be a subset so that whenever $x,y \in C$ there exists a connected subset $A \subseteq C$ so that $x,y \in A$. Prove that $C$ is connected.

Suppose to the contrary, that $C$ is disconnected.
Then there exist non-empty subsets $A$ and $B$ such that $A \cap B = \emptyset$ and $C=A \cup B$. How do we go forward?

Answer 1

First, $A$ and $B$ should be open.

Suppose that we can take $x\in A,\ y\in B.$ Then, by assumption, there exists a connected subset $C'$ of $C$ so that $x,y\in C'.$ Note that $C'\subset A\cup B.$ Can you finish from here?

EDIT: Details for remainder of the proof are in comments below. I'll add them here, as a spoiler:

Consider $A'=A\cap C'$ and $B'=B\cap C'$. These are disjoint, relatively open, non-empty ($x\in A', y\in B'$), and have union $C'$. This forms a disconnection of the connected set $C',$ a contradiction. So, we cannot have arbitrary elements $x,y$ with $x\in A$ and $y\in B.$ Hence, every element of $C$ is contained in either $A$ or $B$, contradicting that both are non-empty.