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 $v_i \ne v_1$ ($2 \le i \le 2n-1$) and a natural number $k$ such that $i+1 \le k \le 2n \land v_i=v_k$. If $k-i$ is odd, then we have an odd circuit smaller in length than $C$: $\left({v_i, \ldots, v_k=v_i}\right)$. If $k-i$ is even, then $\left({v_1, \ldots, v_i, v_{k+1}, \ldots, v_{2n+1}}\right)$ 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 (In every iteration we reduce the length of the circuit and we keep an odd circuit. If the length of the circuit is 3, then it must be a cycle, because there can't be a vertex other than the first, which is also the last, which appears more than once).