Start with one player, say, player A. She faced five opponents, so at least three of those ended the same way - all draws, or all non-draws, say, vs. players B, C and D. Consider three players whose games with player one ended similarly. If even one of their mutual games has the same result.. e.g. B vs D.. the threesome with similar results is found, A-B-D. If none did, then B-C-D is the threesome sought. Either way, the threesome exists.
A follow-up question might be.. how many players are needed for there to necessarily be six players between whom all 15 games ended up the same way? A British fellow called Frank Ramsey looked into those, and far as I know, no one has yet found out the minimum number of elements for a six-element clique.