Suppose contrary. Let $O$ be an arbitrary point on the plane, WLOG assume $O$ is red, and consider the circle with center $O$ and radius $1,\sqrt{3},2,$ respectively. First, consider regular hexagon $A_1,A_2,A_3,A_4,A_5,A_6$ with center $O$ and $\overline{A_1O}=2.$ Easy that the number of red point among $A_i$ is less than $3$ and the number of blue point is less than $5,$ so the only possibility is that $A_i,A_{i+3}$ are red point, WLOG assume $A_1,A_4$ are red and the rest are blue.

Second, consider regular hexagon $B_1,B_2,B_3,B_4,B_5,B_6$ with center $O$ and $\overline{B_1O}=\sqrt{3}$ such that $OB_1$ is angle bisector of $\angle A_1OA_2.$ Let $A_iO\cap \text{unitcircle}\equiv C_i.$ Similar to the above discussion we know $B_i,B_{i+3}$ are red for a certain $i$. We have two case below: Case 1. $B_2,B_5$ are red, that is $B_2B_5$ are perpendicular to $A_1A_4.$ Then one of $C_3,C_5$ must be blue, otherwise $\triangle A_4C_5C_3$ is red, similarly one of $C_2,C_6$ must be blue.

Hence the result follows.

First we prove there must either be a monochromatic equilateral triangle of length $1$ or a monochromatic equilateral triangle of length $\sqrt{3}$. If there exists a monochromatic equilateral triangle of length $1$, then we are done. If there does not, then we can find $2$ points $A,B$ such that $A,B$ are of distinct colours, $AB=1$. Consider isoceles triangle $ABC$ with $AB$ as base, $CA=CB=2$. WLOG $B,C$ are of the same colour(so $A,C$ are of different colours). Let $A$ be red, $C$ be blue. Let $O$ be the midpoint of $AC$. WLOG $O,C$ are the same colour, blue. Now construct equilateral triangles of length $1$ $ \triangle OCD,\triangle OCE$; since there does not exist monochromatic equilateral triangle of length $1$, $D,E$ must be red, thus $\triangle ADE$ is a monochromatic equilateral triangle of length $\sqrt{3}$. Back to the problem, suppose there does not exist a monochromatic equilateral triangle of length $\sqrt{3}$, and a monochromatic rhombus of length $1$. From above there must exist a monochromatic equilateral triangle of length $1$. Consider an equilateral triangle $ABC$ of length $2$, $D,E,F$ are the midpoints of $BC,CA,AB$ respectively, and points $DEF$ are all coloured red. Then $A,B,C$ must all be coloured blue, otherwise there is a monochromatic equilateral rhombus of length $1$ (the rhombi $AEDF, BDFE, CFED$), then $\triangle ABC$ is a monochromatic equilateral triangle of length $2$.