Consider a pyramid having - a square base $ABCD$ with side length $2\sqrt{6}$, and - the vertex $E$ being $6$ units above the center of the base. It follows that the distance between any two points in $\{A,B,C,D,E\}$ is either $2\sqrt{6}$ or $4\sqrt{3}$, and furthermore, the circumsphere of $ABCDE$ has radius exactly $4$. Hence it's possible to choose $A,B,C,D,E$ from our spherical surface. Now Pigeonhole finishes our argument because there must be two points with the same color in $\{A,B,C,D,E\}$.