A line $\ell$ for a set $S$ of points is bisecting if the open halfspaces on either side of $\ell$ contain at most $\frac{|S|}{2}$ points.
Now given point sets $A$ and $B$ in the plane, prove that we can always find a line $\ell$ such that $\ell$ bisects both $A$ and $B$ simultaneously. We can assume that no three points in $A \cup B$ lie on a line and no two points share an x-coordinate.
Note that I am not asking for an algorithm separating $A$ and $B$.
So far I have proven the following claim:
Let $S$ be a point set in the plane containing an even number of points. Prove that, for any point $s \in S$, a bisecting line $\ell$ for $S \setminus \{s\}$ also bisects $S$.
From this claim we can also infer that all separting lines for a point set $P$ with an odd number of points must pass through at least one point $p \in P$. However I am stuck on how to continue. Intuition suggests that one of the lines between points $a \in A$ and $b \in B$ must be simultaneously bisecting (if either of $A$ or $B$ contain an even number of points we can just remove one at random by the above claim). However I cannot seem to prove this is always the case.
Can you give me a hint for the next step?