Consider pairs of diametrically opposed vertices. If such a pair contains two vertices of the same color and that color occurs again someplace, we get our rectangle. Otherwise, assuming there are $k\geq0$ diagonals of one color, the remaining $50-k$ diagonals have vertices colored with two different colors of the remaining $10-k$ ones. There are $\binom{10-k}{2}$ combinations of colors but more than that diagonals ($50-k$), so there must be two pairs with the same colors.

By php there are $10$ vertices of the same colour. Now if among these 10 vertices there is a pair of diametrically opposite vertices (Call them $A$,$C$) of this colour then take another vertices $B$ of this colour.$D$ be the antipode of $B$ So $ABCD$ is a rectangle with $A,B,C$ are of same colour. Otherwise, among the 10 vertices of that same colour all the 10 diametrically opposite vertices of them will be of another 9 colour.So by php 2 vertices will be of same colour.We get a rectangle with vertices of at most 2 colours.