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Who is left out to dry in an oddly matched water fight?

Problem: On a large, flat field, an odd number of odd people are milling about with water pistols. At a signal, everyone turns and squirts the closest person! We may assume, since they are just milling about randomly, there is a unique closest person to squirt. Show that there will always be at least one person left dry.

The following solution was given by Jonathan Schisler and Hieu T. Nguyen:

The two absolute closest people will squirt each other. If any of the remaining people squirt either of these two, one of them will remain dry, since there will be fewer remaining squirts than remaining people. On the other hand, if all the remaining people squirt at each other, then the two closest remaining people will squirt each other. Now consider the remaining people not in this pair. Repeating the argument, we either will eventually produce a dry person, or will continue to split off pairs of people squirting each other, eventually leaving one person the odd one out.

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