Graph containing Closed Walk of Odd Length also contains Odd Cycle

Theorom
Let $G$ be a graph.

Suppose $G$ has a closed walk of odd length. Then $G$ has a odd cycle.

Proof
Let $G = \left({V, E}\right)$ be a graph with closed walk of odd length.

Let us take a closed walk of odd length in $G$, and note it by $C = \left({v_1, \ldots, v_{2n+1}=v_1}\right)$. We can assume without loss of generality that $C$ is a circuit because Closed Walk of Odd Length contains an Odd Circuit.

Assume $G$ has no odd cycles. Then $C$ is not a cycle. Hence, there exist a vertex $v_i \ne v_1$ ($2 \le i \le 2n$) and a natural number $k$ such that $i+1 \le k \le 2n \land v_i=v_k$. If $k$ is odd, then we have an odd circuit smaller in length than $C$: $\left({v_i, \ldots, v_k=v_i}\right)$. If $k$ is even, then $\left({v_1, \ldots, v_{i-1}, v_i, v_{k+1}, \ldots, v_{2n+1}}\right)$ is an odd circuit smaller in length than $C$. We can take the new odd circuit as $C$, and the same arguments hold. In this way we can reduce the length of the odd circuit until we have an odd cycle.