If we color the plain with three different colours, then there will always be an equilateral triangle which has three vertices of the same colour.
I have proved it for two colours but I just can't solve this one.
This needs van der Waerden's Theorem. There exists $N(r,k)$ such that in any line of $N$ equally spaced points which are each coloured with one of $r$ colours, we can always find $k$ equally spaced points which are monochromatic.
Let $N=N(2,3)$ and $M=N(3,N+1)$. Then if we take any $M$ equally spaced collinear points in the 3-coloured plane, we can find $N$ equally spaced monochromatic points along a line segment $L$. Suppose they are coloured 1.
That gives us a large triangle of equally spaced points in the equilateral triangle with side $L$. If any of them is also coloured 1, then we have a monchromatic equilateral triangle with two points in $L$. So assume they are all coloured 2 or 3.
The next line up in the triangle has $N$ points, so we can find 3 of them which are equally spaced and monochromatic. Suppose they are coloured 2. Call them $A,B,C$. Further up in the large triangle we have $D$ so that $ABD$ is equilateral. Now if it is not monochromatic $D$ must be coloured 3. Similarly, the point $E$, with $BCE$ equilateral must also be coloured 3. But now the point $F$ further up in the large triangle makes $DEF$ equilateral.
But however it is coloured, $F$ must be in a monochromatic equilateral triangle. The other two vertices of $DEF$ are coloured 2, the other two vertices of $BDF$ are coloured 3, and there are two vertices in $L$ forming an equilateral triangle with $F$ which are coloured 1.