Coloring $\mathbb R^3$ with 3 colors always gives us a color with all distances. I'm having some issues reasoning through a particular problem. Consider a
problem where I color all points in $\mathbb{R}^3$ with the colors red, green,
blue. I want to show that every positive real represents the distance between
two points of one of these colors (one color attains all distances). I have a
feeling that this is to some extent an "assume for sake of contradiction"
kind of proof. Let us say that there is no color that attains all distances,
i.e. red does not attain $d_{r},$ blue does not attain $d_{b},$ and green does
not attain $d_{g}.$ $d_{r}$ represents the distance between blue or green
points, $d_{b}$ represents the distance between red or green points, and
$d_{g}$ represents the distance between red or blue points. However, once I
reach this point, I am running out of ideas of where to move with this proof.
Any assistance would be greatly appreciated.

Suppose red, blue, green don't attain $d_r$, $d_b$, $d_g$ respectively, with $d_r \ge d_b \ge d_g$.  I'll leave it to you to handle the case where there is no red point.  Otherwise, take a red point $p_r$.  Then the sphere $S$ of radius $d_r$ centred there contains only blue and green.  If it's all green,
green  would attain $d_g$ for any two points at distance $d_g$ on $S$, so there is at least one blue point $p_b$ on $S$.  The intersection of $S$ with the sphere of radius $d_b$ centred at $p_b$ is an all-green circle of radius $ y > d_b/2 \ge d_g/2$ so this circle contains
two points at distance $d_g$, contradiction. 
