Solution:

Consider the lattice points (points with integer coordinates) on the lines $y=0$ and $y=1$, other than $(0,0)$ and $(0,1)$. If one of them, say $A=(p, 1)$, is coloured green, then we have a right-angled triangle with $(0,0)$, $(0,1)$ and $A$ as vertices, all having different colours. (See Figures 1 and 2.)

If not, the lattice points on $y=0$ and $y=1$ are all red or blue. We consider three different cases.

Case 1. Suppose a point $B=(c, 0)$ is blue. Consider a green point $D=(p, q)$ in the plane. Suppose $p \neq 0$. If its projection $(p, 0)$ on the $x$-axis is red, then $(p, q)$, $(p, 0)$ and $(c, 0)$ are the vertices of a required type of right-angled triangle. If $(p, 0)$ is blue, then we can consider the triangle whose vertices are $(0,0)$, $(p, 0)$ and $(p, q)$. If $p=0$, then the points $D$, $(0,0)$ and $(c, 0)$ will work. (Figure 3.)

Case 2. A point $D=(c, 1)$, on the line $y=1$, is red. A similar argument works in this case.

Fig-4

Case 3. Suppose all the lattice points on the line $y=0$ are red and all on the line $y=1$ are blue points. Consider a green point $E=(p, q)$, where $q \neq 0$ and $q \neq 1$. (See Figure 4.) Consider an isosceles right-angled triangle $E K M$ with $\angle E=90^{\circ}$ such that the hypotenuse $K M$ is a part of the $x$-axis. Let $E M$ intersect $y=$ in $L$. Then $K$ is a red point and $L$ is a blue point. Hence $E K L$ is a desired triangle.