If there are three points, by pigeonhole there must be two of the same color making a line segment.

I don't see how that helps. My solution is this: Yes, if it is possible we will work in $\mathbb{R}^2$ we choose all the circles with center in the origin If a circle has a rational radius, we paint all its points red. If it has an irrational radius we paint them blue [asy][asy] dot((0,0), red); draw((0,-10)--(0,10)); draw((-10,0)--(10,0)); draw(circle((0,0),5), red+linewidth(1)); draw(circle((0,0),3.14), blue+linewidth(1)); [/asy][/asy] We will check that it meets: Suppose that $\exists$ a segment $AB$ such that all its points are of the same color. We know that points that are a distance $x$ from the origin form a circle, and since a segment cannot lie completely on a circle $\Rightarrow$ $\exists$ 2 points $\in AB$ such that they have the maximum and minimum distances, respectively. Let these distances be $r_1$ and $r_2$ respectively It is easy to notice that $\exists$ a point in the segment such that its distance to the origin is a, $\forall$ a $\in [ r_2,r_1 ]$ By the density of rationals and irrationals $\exists$ a point $\in AB$ such that its distance to the origin is rational and $\exists$ a point $\in AB$ such that its distance from the origin is irrational. $\Rightarrow$ There is always at least one red and one blue dot in each segment. $(\Rightarrow \Leftarrow)_\blacksquare$