Thirteen men are standing in a field, each with a pistol. The distance between each pair of men is unique, i.e., no two pairs are the same distance apart. At 12:00 noon sharp, each man shoots at and kills his nearest neighbour. Will there always be a survivor?

There WILL always be at least one survior.

To see this, draw a graph where each vertex represents a man, and draw a directed edge from vertex u to vertex v if u shoots at v.

Now, suppose that every man is shot by some other man. That means that at each vertex there is at least one edge coming in, and exactly one edge going out.

Starting from any vertex, pick any edge coming in and follow it back to the preceding vertex. From that vertex, again pick any edge coming in and follow it back. Since there is a finite number of vertices, this process will eventually end up with vertex that we have already encountered. In fact, it has to be the original vertex, since if it ended up with one of the other vertices, then that vertex would have two outgoing edges, implying that that man shot at two neighbours. So this procedure produces a loop.

If the loop doesn't contain all of the vertices, then we can repeat the procedure starting with one of the remaining vertices. Again, pick any edge coming in and follow it back, and repeat that process. We won't encounter any vertices of the first loop, since that would yield a vertex with two outgoing edges, nor will we encounter any vertex in the second path other than the first, for the same reason.

By continuing this process, we can produce a collection of loops that contain all of the vertices.

Now we claim that none of these loops has more than two edges. Suppose there
is a loop u_{1}, u_{2}, ... u_{n}, u_{1},
with n > 2. Because of the way the loop is constructed, each man is shot at by the
next man in the loop. Because each man shot at his nearest neighbour, and
using the fact that the distances are unique, we deduce that the
distance between each pair of members in the loop is strictly increasing,
i.e.:

d(u_{1},u_{2}) < d(u_{2},u_{3}) < ... < d(u_{n},u_{1})

However, this is impossible: u_{1} fired at u_{n}, so
d(u_{1},u_{2}) > d(u_{1},u_{n}).

As a result, all of the loops must be of size two, which implies that the total number of men is an even number and cannot be 13. This contradiction establishes the result.