Let be a topological group acting on continuously and properly discontinuously.

Assume that does not act freely, i.e. there exist with such that for some . Then for any neighbourhood of , , violating proper discontinuity. Thus acts freely.

Now consider the orbit of a point with its subspace topology and the corresponding orbit map .

Assume there exists with not open in . Let be an open neighbourhood of in . Since is a homeomorphism, is open in , and thus is open in , so at least one distinct point is contained in . Then , violating proper discontinuity. Therefore must be discrete.

Now clearly the orbit map is continuous and bijective (injectivity by freeness, surjectivity by construction). Thus every singleton in is the preïmage of a singleton in and is therefore open. Therefore is discrete, and is a homeomorphism, since the inverse is continuous as a map between discrete spaces.