I am trying to prove what's on the title. I have been working on it for some time already and the problem I have is that I can't seem to prove that the cycle I get at the end is of odd length.
Here are the conclusions I reached, which I am not sure at all are correct:
If G contains a closed walk of odd length (let's say a u-v walk), then it contains 2 u-v walks, one of even length and another one of odd length so that when added up they give an odd number. Then for every of these u-v walks, we can obtain a u-v path by removing all the repeated fragments of the walk.
We are left now with 2 paths. These two paths might form a cycle or not.
In the case they do form a cycle, we are done (however I know nothing about the length of such cycle).
In the case they do not form a cycle, remove all xy edges s.t. xy belongs to both paths. We will remove an even number of edges before we reach the cycle; but then again, I can't tell the parity of the length).
Thanks a lot!