The approach taken by Kelley in his classic bookGeneral Topologyis to prove 4. first, because that is comparatively easy. If denotes the equivalence class of the constant sequence (x,x,x,...), and similarly for , then it's more or less obvious that . So the embedding is isometric. To see that its range is dense, suppose that z is an element of and let be a sequence in the equivalence class z. Then the sequence converges to z in the metric given by p. (That's not difficult, once you've got your head around the idea of that construction.)

Let be the isometric embedding . To prove that is complete (problem 3.), I'll quote Kelley's argument practically word for word (it's short but clever). Observe first that it is sufficient to show that each Cauchy sequence in f(X) converges to a point of (because we have already proved that f(X) is dense in ). But each Cauchy sequence in f(X) is of the form , for some sequence in X. Let be the equivalence class of the sequence . Then as n→∞. (Again, this is almost immediate, once you see what's going on here.)

Finally, for 5., suppose that g:X→Y is another isometric map with dense range. Then the map is an isometry between dense subspaces of and Y, which extends by continuity to an isometry from the whole of to the whole of Y (that's Kelley's argument again).