Prove: For any 2 coloring of 2-space, one of the color classes contains points at all distances

Question

We color the 2D-plane either red or blue at every point. Prove that one of the sets (either red or blue), contains two points at distance $D$, for every positive real number $D$.

Answer 1

Hint Pick a red point and consider a circle of radius $D$. What can be said about all the points on the circle? Pick a point on the circle we just made and consider a circle of radius $D$ there. What can be said about all the points on the circle? Note that these two circles intersect.

(This may not be the easiest method but it is the first thing that sprang to mind.)

Answer 2

Let $R,B\subseteq\mathbb R_+$ be the sets of distances covered by red and blue, respectively. Assume both miss some distance, say $0<r\notin R$ and $0<b\notin B$. Then there exists a blue point $A$ (or the plane would be all red and trivially $R=\mathbb R_+$). If $b\ge \frac r2$ the circle of radius $b$ around $A$ contains two points $B,C$ of distance $r$, hence these cannot both be red, hence one of $AB$, $AC$ shows that $b\in B$. We conclude $b<\frac r2$. By the same argument $r<\frac b2$, contradiction.