Let $W$ be a closed walk of length at least $1$ that does not contain a cycle. Prove that some edge of $W$ repeats immediately (once in each direction).
How do I prove this? Can I do a contradiction and say "Assume that no edge of $W$ repeats. Therefore since $W$ is a closed walk with no repeated edge it is a cycle or it contains a cycle (since repeated vertices are allowed) which is a contradiction."Any hints or solutions would be of great help.
The proof is trivial if $W$ is length $2$...
so then for $W$ with length greater than $2$: If we have $W:v_1v_2\ldots v_nv_1$, consider a subwalk $W_1:v_1v_2\ldots v_kv_1$ in which $v_1$ is not an internal vertex (it could be that $k=n$). Then we must have $v_k=v_2$ for else we have a $v_2-v_k$ walk not containing $v_1$, and such a walk must contain a $v_2-v_k$ path not containing $v_1$. If we take such a path together with the path $v_kv_1v_2$ we have a cycle (contained in $W_1$, and therefore also in $W$), which is a contradiction. So we have $W_1:v_1v_2\ldots v_2v_1$.
Now we can find a subwalk $W_2:v_2v_3 \ldots v_jv_2$ contained in $W_1$, so that $v_2$ is not an internal vertex of $W_2$, and by a similar argument we can show that we must have $v_j=v_3$.
Continuing this process, and since $W$ is finite, we will eventually have a subwalk $W_i:v_iv_{i+1}v_i$ of $W$ which contains an edge that is immediately repeated in the opposite direction.