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@above shouldn’t you prove no subdivision of $K_5$ and $K_{3,3}$ exist, as opposed to just $K_5$ and $K_{3,3}$?

CT17 wrote: Solved with Eyed, amuthup, ETS1331, AI216 We claim the answer is no. Let the distance we are considering be $d$. Draw a graph where two points are connected iff they are distance $d$ apart. Clearly this graph contains no $K_5$ or $K_{3,3}$, so it is planar by Kuratowski's theorem. But then it has at most $3\cdot 330 - 6 = 984$ edges, so we're done.

IMPORTANT SPOILER