Assume the contrary. Let the colors be "read" and "blue". Let the line be the real number line. So there shall be now three numbers $x,y,z$ having the same color and satisfying $x+z=2y$ Take two arbitrary points $2x$, $2y$ having the same color, say "red". Then $x+y, 4y-2x, 4x-2y$ must be colored "blue". But then we have $(4y-2x)+(4x-2y)=2(x+y)$. Contradiction! Hence the result.

Suppose for the sake of contradiction that there aren't three points $A,B,C$ of the same colour such that $AB=BC$. Let's say that the colours are red and blue.Let $M,N$ be two red points and let $P$ be the midpoint of segment $[MN]$.Then the point $P$ is blue.Let $M'$ be the symmetric of point $M$ with respect to $N$ and let $N'$ be the symmetric of point $N$ with respect to $M$.Then the points $M',N'$ are both coloured blue and $P$ is the midpoint of segment $[M'N']$,but $P$ is blue,contradiction!This ends our proof.