$55$ gangsters shoot the nearest gangster to them, where all the distances between them are different. Prove that at least one gangster will survive. My thinking was that if gangster A shoots gangster B, then gangster B will also shoot gangster A since they are the closest together, forming a pair. Since 55 is odd, then one must survive since 55 is 1 mod 2.

The pair of gangsters with lowest pairwise distance will shoot each other. If some other gangster shoots any of those two, then there will be at most $52$ bullets aimed at the remaining $53$ gangsters, and hence at least one will survive.

If no other gangster shoots any of those two, the problem is reduced to the case of $53$ gangsters and we proceed by induction. At this point, it boils down to checking that the case for $3$ gangsters always ends up with one alive.