Assume, if possible, that the result is false so every equilateral triangle in the plane has two vertices of different colours. Consider a regular hexagon $ABCDEF$ with centre $S$.

Assume $S$ is coloured black. One of the vertices of the triangle $BDF$ must be black; assume it is $B$. Considering the equilateral triangles $SBA$ and $SBC$, then $A$ and $C'$ are both white. From the triangle $ACE$, then $E$ is black and from $SEF$ the vertex $F$ is white. If $G$ is the point of intersection of $BA$ and $EF$, then we get a contradiction for the colouring of $G$. If it is white (black) then the triangle $GAF$ ($GBE$) is an equilateral triangle with all the vertices of the same colour. So we conclude that there is an equilateral triangle whose vertices do have the same colour.