For an odd degree node . Let $v$ be an odd degree node. Consider the longest walk starting at $v$ that does not repeat any edges (though it may omit some). Let $w$ be the final node of the walk . Show that $v$ $\neq$ $w$. **My approach** :By well ordering principal,consider a walk $S$ of minimal length. By contradiction let $v$ $=$ $w$. Removing an edge $e$ from the end-point gives a shorter walk. This contradicts our assumption of minimality .

I can't follow your argument at all. You seem to be proving that a _shortest_ walk cannot begin and end at the same node (which is at least doubtful -- how about the empty walk?), but the question was about a _longest_ one.

Instead, if the walk both begins and ends at $v$, then the walk have used an _even_ number of edges that touch $v$ -- namely, you _leave_ $v$ the same number of times you _come back_ to it. Since $v$ had odd degree, this means there must be at least one edge left unused that you could continue the walk along -- so it certainly wasn't the _longest_ walk that doesn't repeat edges.