My solution: We need not have the points to be named as such. For convenience, we name them as we proceed. Consider the five line segments from a point $A$ to the other points. By pigeonhole principle, there exists a tetrahedron $ABCD$ with $AB, AC, AD$ of same colour, say black. Suppose we can't find a monochromatic triangle. Since $ABC, ACD$ are non-monochromatic, $BC, CD$ are coloured white. Now, consider $BD$. If it is black, then $ABD$ is monochromatic and if it is white, $BCD$ is monochromatic and so, we need to have at least one triangle with edges of same colour.