Suppose I have an even degree undirected graph G. Given an algorithm to find a tour in G FINDTOUR(G,s), where s is the starting vertex, prove by induction on the length of the walk that the only place the algorithm will get stuck is at s.
FINDTOUR(G,s): start by walking from a vertex s, at each step choose an untraversed edge incident to the current vertex until you get stuck.
Proof by induction on n, the length of the walk in an even degree graph G=(V,E)
Base Case: for n=0 the claim is trivially true as there is no tour.
Inductive Hypothesis: Suppose that for a walk of length $n = k \geq 0$ the algorithm gets stuck at the starting vertex.
*Inductive Step: This is where I'm unsure of how to proceed.
Intuitively the claim makes sense because if the graph has an even degree then every vertex $v \neq s$ will have an odd number of untraversed incident edges to it.