We argue by contradiction. Assume such a coloring exists.
Start at any point in the plane P. Choose a ray emanating from P. There is an equilateral triangle (of side length 1) with one vertex at P where the ray goes through the interior as an axis of symmetry. Let the two other vertices be called Q and R. There is another distinct equilateral triangle with vertices Q,R,S. Note that S lies on the ray we chose. Now the distance from P to S can be easily calculated as the square root of 3.
The color of P and S must be the same. This is because the colors of P,Q,R must all be different, and so must the colors of Q,R,S. Since there are only three colors, P and S must be the color missed by Q and R.
So, this shows that every point of distance square root of 3 from P must be colored the same. But it is easy to show that there are two points on this circle of distance 1 apart. This is a contradiction.