If each point is joined to every other point then this makes (6 x 5)/2 = 15 line segments. All the line segments are coloured either red or green.
Joel of ACS (Barker), Singapore proved that there must then be a triangle all of whose sides are the same colour. This is his solution:
Label the points from A to F. Out of the line segments connected to A, at least 3 have to be the same colour, (let's say AB, AC and AD are red). Then BC has to be green, because otherwise ABC will be all red. Ditto for CD and BD. So BCD will be all green.