Problem 914

16.4. Suppose, contrary to the statement of the problem, that any plane intersects no more than 3 sets. Let's choose points $A, B, C, D, E$ from different sets. Then no 4 of them lie in the same plane, and, consequently, no 3 lie on the same line. Further, some plane passes through any 3 of them, relative to which the other 2 points are located in different half-spaces (this property is possessed by at least one of the planes $A B C$, $A B D$, $A B E$). Let this plane pass through points $A, B, C$. The point $F$ of intersection with it of the line $D E$ belongs to one of the sets containing points $A, B, C$, for example, the set containing point $A$. Therefore, the plane passing through points $D, E$, $F, B$ intersects no fewer than four sets. The resulting contradiction proves the validity of the statement of the problem.