I am looking for a non-compact connected space $X$ such that for any two disjoint closed $A,B\subseteq X$ there exists a proper closed connected $C\subseteq X$ such that $A\cup B\subseteq C$.
I would like the space to be normal if possible.
An example of a compact connected space with this property is the circle $S^1$: If $A$ and $B$ are disjoint closed sets in $S^1$ then $A\cup B\subsetneq S^1$ since $S^1$ is connected, so $S^1 \setminus (A\cup B)$ contains an open segment $(a,b)$. Then $S^1 \setminus (a,b)$ is as desired.
Note that the reals do not have this property.
Here's an even easier way to show $\mathbb R^n$ has this for all $n\geq 2$, basically using your $S^1$ argument.
The complement of $A\cup B$ is open and non-empty, so contains a pair of small open balls $U\subsetneq V$. Now take $C=U^c$. This is closed and contains $A\cup B$ by construction, and is (path) connected when $n\geq 2$.