Five points contain a convex quadrilateral

Idea. Split according to the number of vertices of the convex hull.

Hull with $4$ or $5$ vertices. Then any $4$ hull vertices are in convex position, and we are done.

Hull a triangle $ABC$. The other two points $D,E$ are inside the triangle. The line $DE$ does not pass through $A,B,C$, so each of $A,B,C$ is on one side of it. By pigeonhole, two of them, say $B$ and $C$, lie on the same side of $DE$.

We claim $B,C,E,D$ are in convex position.
- $B$ and $C$ are vertices of the hull, so they cannot lie in the convex hull of other points.
- $D$ is not in triangle $BCE$: this triangle touches the line $DE$ only at $E$, while $D\ne E$ is on that line. Symmetrically $E$ is not in triangle $BCD$.

So the quadrilateral with vertices $B,C,E,D$ is convex. ∎

Remark. This is the first case of the Erdős–Szekeres "Happy Ending" problem.

Lesson. Case on the convex hull, then use pigeonhole on which side of a well-chosen line the vertices lie.