By # 2471, if all the points with distance of $2$ have the same color, then all points have the same color.

If all points have the same color, the conclusion holds.

Otherwise, if there exist two points, $A$ and $B$ of different colors and is separated by $2$, then its middle point $O$ must have the same color as one of them. Let's assume it is $A$. Then, construct two equilateral triangles $\triangle{AOC}$ and $\triangle{AOD}$. If one of $C$ and $D$ has the same color as $A$ and $O$, then we find a qualified triangle with side length of $1$. Otherwise, $\triangle{BCD}$ is a qualified triangle with side length $\sqrt{3}$.