Lemma 5.4.2. If there is a walk from a vertex u to a vertex v in a graph, then there is a path from u to v.
Proof. Since there is a walk from u to v, there must, by the Well-ordering Principle, be a minimum length walk from u to v. If the minimum length is zero or one, this minimum length walk is itself a path from u to v. Otherwise, there is a minimum length walk v0,v1......vk. from u=v0 to v=vk where k>=2. We claim this walk must be a path. To prove the claim, suppose to the contrary that the walk is not a path; that is, some vertex on the walk occurs twice. This means that there are integers i; j such that 0<=i<=j<=k with vi=vj. Then deleting the subsequence v(i+1).....vj yields a strictly shorter walk v0,v1......vi,vj+1,vj+2....vk from u to v, contradicting the minimality of the given walk.
I can't figure out how the Well ordering Principle comes in. The Well ordering Principle is about "Every nonempty set of nonnegative integers has a smallest element." Since a walk is a sequence of vertices and edges, how do they relate set of nonnegative integers to a walk?