Proving that if a graph $G = (V,E)$ has an odd closed walk (i.e. odd number of edges) , then $G$ has an odd cycle. Hi I am new in graph theory and I would sincerely appreciate your help with my proof. This is not for a homework, I am just practicing for one of my courses.
So, my approach to prove this statement was by induction on the length of the closed walks. I already have the base case. Now, assume that the statement is true for any closed walk $(v_1, v_2, ..., v_k)$ where k < n and k is odd. Assume we have a closed walk $A = (v_1, ..., v_n = v_1)$ where n is odd. So if we do not have any repeated vertices other than $v_1$ and $v_n$ then we are done and $A$ is an odd cycle. If not, then we have some $v_i$  such that $A$ can be seen as $(v_1, ..., a, v_i, ..., v_i, b,..., v_n  = v_1)$. So we have one closed walk $ W = (v_i, ..., v_i)$ that has length $<n$ and we can apply induction hypothesis on that walk. This is the point where I do not know how to prove. I would like to say that the sequence $V = (v_1, ...., a, b, ...., v_n = v_1)$ is a closed walk. But how do I even know that it is a walk ? Also how do I know that it is disjoint from the other walk? I have seen that this technique is applied in some proofs of graph theory but I am not sure why V would be a closed walk.
Again any help would be greatly appreciated and, as I said, this is not a problem from an assignment.
 A: Suppose that $i<j$, and $v_i=v_j$. Then $v_iv_{i+1}\ldots v_{j-1}v_j$ is a closed walk, and $v_jv_{j+1}\ldots v_{n-1}v_1\ldots v_i$ is also a closed walk. If $\ell_1$ is the length of the first walk and $\ell_2$ the length of the second walk, then $\ell_1+\ell_2=n$, and since $n$ is odd, one of $\ell_1$ and $\ell_2$ must be odd and the other even. Pick the one of odd length, and apply your induction hypothesis.
If I understand your argument correctly, you want to cut out $v_{i+1}\ldots v_j$ and use the path $v_1\ldots v_iv_{j+1}\ldots v_n$. That’s my second path, just listed from a different starting point, and there’s no guarantee that its length is odd: $\ell_2$ might be even, in which case it’s the other path, the one through the deleted part, that you want to use.
A: 
If not, then we have some $v_i$  such that $A$ can be seen as $(v_1, ..., a, v_i, a, ..., v_i, ..., v_n  = v_1)$

Why not write $A$ starting and ending at the duplicated vertex, which I'll call $v$ for ease of notation, and the other vertices $a_i$ and $b_i$:
$$A=va_1\dots a_mvb_1\dots b_n(v)$$
Since $A$ has odd length, $m+n+2$ is odd, i.e. $m+n$ is odd, i.e. exactly one of $m$ or $n$ is even. Furthermore, both $m$ and $n$ must obviously be at least $1$. Without loss of generality take $m$ to be even; it must be at least $2$. Then $va_1\dots a_m(v)$ has strictly fewer vertices than $A$ and has an odd length at least $3$, whereupon we can apply the induction hypothesis.
