By the pigeonhole principle, we have $5$ points in the same colour, wlog say that colour is red. The other four points are therefore blue. Consider those $5$ points and consider those cases: 1. Suppose we have $(1,1)$ coloured red, then we have two subcases: we have red at the edge (not corner) or we have red in one of the corners. For both cases, eliminate directly bad places for those red points and obtain trivial contradiction. 2. Hence, we have $(1,1)$ blue. Now we need to place $5$ red points at the edge. Let us call $4$ corners ($(0,0)$,$(2,0)$,$(0,2)$ and $(2,2)$) one region and the other $4$ points ($(0,1)$, $(1,0)$, $(1,2)$ and $(2,1)$) as another region. By pigeonhole principle, we must have $3$ points in the same region. Those three red points form a right isosceles triangle. We are done.