Let us look at this version of the original problem in which the number of sets is increased by one:
We have 11 points on the plane. We want to show that we can divide them into three sets of three points and one set of two points such that the intersection of the convex hulls of all four sets is not empty.
We take 10 of the points and apply Tverberg’s theorem.
It gives us four disjoint nonempty sets which together contain all 10 points. The intersection of the convex hulls of these sets is nonempty. Let on of the points in the intersection of the convex hulls be x.
If any of the four sets are bigger than 3 then by Caratheodory’s theorem there is a three point subset that contains x. We take that as the new set in place of the four point set and add the additional point to a set less than three. We repeat this process untill all of the sets are three or less. Then we add the 11th point to a set with two points or less. The result is 11 points divided into four sets of three or less. The only way to do this is three sets of three points and one set of two points which gives the desired theorem.