Proof of the theorem on friends and strangers

Proof

We look at the edges at one vertex and use the pigeonhole principle.

1. Colour each edge of the complete graph $K_6$ red or blue. Fix a vertex $v$.
2. The vertex $v$ is adjacent to the other $5$ vertices, so it lies on $5$ edges.
3. Suppose that at most $2$ of these edges were red and at most $2$ were blue. Then there would be at most $4$ edges, a contradiction. This is the pigeonhole principle for $5$ edges and $2$ colours.
4. So at least $3$ of these edges have one colour. By symmetry of the two colours, we may call this colour blue.
5. Let $a, b, c$ be three distinct vertices such that $va$, $vb$ and $vc$ are blue.
6. Case 1: one of the edges $ab$, $bc$, $ca$ is blue, say $ab$. Then the edges $va$, $vb$ and $ab$ between $v, a, b$ are all blue.
7. Case 2: none of $ab$, $bc$, $ca$ is blue. Then all three are red, and the three edges between $a, b, c$ are all red.
8. In both cases there are three vertices such that the three edges between them have the same colour.