Coloring of $\mathbb{R}^3$ into 3 colors Every point of three-dimensional space is colored red, green, or blue. Prove that one of the colors attains all distances, meaning that any positive real number represents the distance between two points of this color.
My proof: Suppose by contradiction that $\exists \delta>0$ such that for any $x,y\in \mathbb{R}^3$ with $d(x,y)=\delta$ points $x$ and $y$ have different colors. Let's consider the tetrahedron in $\mathbb{R}^3$ with base $A_1A_2A_3$ and upper vertex $A_4$. Suppose $A_1$ is colored into red,$A_2$ is colored into green then $A_3$ is colored into blue. Then $A_4$ should be colored into one of the colors blue, red, or green but due to $d(A_4,A_1)=d(A_4,A_2)=d(A_4,A_3)=\delta$ the coloring of point $A_4$ to red,blue, or green is impossible.
EDIT: Let's prove that RED attains all distances.Suppose by contradiction that $\exists \delta>0$ such that for any $x,y\in \mathbb{R}^3$ with $d(x,y)=\delta$ points $x$ and $y$ not both RED. Let's consider the regular tetrahedron in $\mathbb{R}^3$ with base $A_1A_2A_3$ and upper vertex $A_4$. WLOG suppose $A_1$ is colored into red,$\ A_2$ is colored into green then $A_3$ is colored into blue. Then $A_4$ should be colored into one of the colors blue, red, or green but due to $d(A_4,A_1)=d(A_4,A_2)=d(A_4,A_3)=\delta$ but the coloring of point $A_4$ to red,blue, or green is impossible. Here we get contradiction.
Right?
 A: You have proven that for any distance $\delta$, we can find two points at that distance whose colors match.  What you haven't proven (and what is requested) is that we can always find two points at that distance that are both a particular color (red, for example).  Your construction leaves open the possibility that for (say) $\delta = 1$ all pairs thus constructed are blue, while for $\delta = 2$ all pairs thus constructed are red.
EDIT:  I'm not entirely sure what the logic is in your revised proof;  the concluding sentences are rather unclear to me.  However, it appears to assume that the vertices of any tetrahedron include all three colors, but this is not necessarily the case.
A: You're on the right track, but your proof could do with a bit of tidying to make it clearer.
Suppose for contradiction that $\exists \delta>0$ such that for any $x,y\in \mathbb{R}^3$ with $d(x,y)=\delta$ points $x$ and $y$ have different colors. 
Consider the regular tetrahedron in $\mathbb{R}^3$ with all side lengths equal to $\delta$, and label the base $A_1A_2A_3$ and upper vertex $A_4$. Without loss of generality, take $A_1$ coloured red. Then $A_2$ must be differently coloured, say it is green. Then $A_3$ must be coloured differently to red and green, so say it is blue.  But now, $A_4$ cannot be coloured red, green or blue, and we reach our desired contradiction.
The two important changes I've made are to specify the tetrahedron has side length $\delta$, and explain why we took $A_1$ red, $A_2$ green and so on.
EDIT: Michael Seifert points out more important issues with your proof which I missed.
A: Suppose there is no color which attains all distances. This means there exists a distance $r$ not attained by any two red points, a distance $b$ not attained by any two blue points, and same for $g$ with green points. WLOG $r\ge b\ge g$. 
Let $x$ be any red point*. Let $S_x$ be the (surface of the) sphere of radius $r$ centered at $x$. Every point in $S_x$ must be blue or green.
Let $y$ be any blue point in $S_x$**. Let $S_y$ be the sphere of radius $b$ centered at $y$. Since $b\le r$, the intersection of $S_x$ and $S_y$ is a circle. Every point in that circle must be green. 
After drawing a picture, you should be able to convince yourself that the diameter of this circle is at least $b$; since $b\ge g$, there will exist two points on the circle whose distance is $g$, which contradicts the supposed fact that no two green points are at distance $g$. 
*Here we are assuming that there exists a red point. What if there are no red points? This is left as an exercise to the reader. 
**What do you do if no blue point exists?
