A secondary school problem: Prove that for any colouring of a plane with three colours there exists a segment of length 1 with edges of same colour.
My attempt to prove. Take any point A on the plane and let it have colour 1. Draw a circle of radius 1 with the centre at A, where all points on the circumference must have a colour different from 1, otherwise the segment is found. Take a regular hexagon inscribed into the circle. Its sides have length 1 and the vertices have colours 2 and 3, while no adjacent vertices must have same colour. So vertices $B_1$, $B_3$, $B_5$ have colour 2, vertices $B_2$, $B_4$, $B_6$ have colour 3 and no contradiction!
I tried spreading the triangles making a triangular grid covering the plane. It doesn't help either, as any time the colour for a new vertex can be found.