Prove or disprove if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ trail such that every edge of $G$ lies on one of these trails. Let $G$ be a connected graph with exactly 2 odd vertices $u$ and $v$ such that $deg(u) \geq 3$ and $deg(v) \geq 3$. Prove or disprove if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ trail in $G$ such that every edge of $G$ lies on one of these trails.
I know that if if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ path in $G$. But the rest bother me, I feel like they want to say $G$ contains an Eulerian trail. I tried to find a counter example, but got no luck.
I don't know if this is correct, but let's try it. Because $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ paths in $G$ called $p_1$ and $p_2$. But if one of these trail , says $p_1$, contain every edge of $G$ then it also contain edges in $p_2$, so $p_1$ and $p_2$ can't be edge-disjoint paths. So this is false because it contradict itself, right?

I'm going to disagree with the answers so far and say that the result is false.
Not only is it false, but no graph satisfying all the conditions can possibly have two such trails.
Below is an expertly drawn 2-edge connected graph with exactly two odd vertices $u,v$. It's clearly not possible to cover all the edges with exactly two edge-disjoint trails from $u$ to $v$. You can do it with one trail or three trails, but not two trails.

But why is it never true?
Say that we have a trail from $u$ to $v$. Remove it. All the vertices will now have even degree, so any trail that uses all the edges must end up where it started and so can't be a trail from $u$ to $v$.
