Graph containing Closed Walk of Odd Length also contains Odd Cycle

Theorem
Let $G$ be a graph.

Let $G$ have a closed walk of odd length.

Then $G$ has an odd cycle.

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

From Closed Walk of Odd Length contains Odd Circuit, such a walk contains a circuit whose length is odd.

Let $C_1 = \left({v_1, \ldots, v_{2n+1} = v_1}\right)$ be such a circuit.

Aiming for a contradiction, suppose $G$ has no odd cycles.

Then $C_1$ is not a cycle.

Hence, there exist a vertex $v_i$ where $2 \le i \le 2n-1$ and an integer $k$ such that $i+1 \le k \le 2n$ and $v_i = v_k$.

If $k-i$ is odd, then we have an odd circuit $\left({v_i, \ldots, v_k = v_i}\right)$ smaller in length than $C_1$.

If $k-i$ is even, then $\left({v_1, \ldots, v_i, v_{k+1}, \ldots, v_{2n+1}}\right)$ is a circuit whose length is odd smaller in length than $C_2$.

This new odd length circuit is named $C_2$, and the same argument is applied as to $C_1$.

Thus at each step a circuit whose length is odd is reduced.

At the $n$th step for some $n \in \N$, either:
 * $(1): \quad C_n$ is a cycle, which contradiction the supposition that $G$ has no odd cycles

or:
 * $(2): \quad C_n$ is a circuit whose length is $3$.

But from Circuit of Length 3 is Cycle, $C_n$ is a cycle, which by definition has odd length.

From this contradiction it follows that $G$ has at least one odd cycle.