 Solution

Fix $w\in W$. It has five incident edges, so three of them have the same color by the pigeonhole principle. Suppose they are red, with other endpoints $a,b,c$. Any existing edge among $a,b,c$ must be blue, or it would complete a red triangle with $w$.

If all three edges among $a,b,c$ existed, they would therefore form a blue triangle. Thus their triple must contain the unique missing edge $uv$. In particular, both $u$ and $v$ belong to the three red neighbors, so $wu$ and $wv$ are both red. The identical argument works if the chosen three edges are blue.

Therefore $wu$ and $wv$ have the same color for every $w\in W$. The missing edge is crucial: it is the only way the monochromatic neighbor triangle can fail to be present.