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?

As you said, if $G$ is 2-edge-connected then there exist 2 edges disjoint $u−v$ paths in $G$.
Let's suppose that those path are maximal, that is, you can't extend those paths with other edges, without intersecting them (a pair of maximal paths always exist). Suppose also that they do not cover all the edges in $G$.
That means that there exists a node in one of those paths whose edges are not totally covered by the two paths. It's easy to see that the number of such edges uncovered is even.
If you now delete all the edges covered by the two paths, you obtain a graph where all nodes have even degrees, meaning every node is included in an Eulerian cicle in his connected component. Take now a node in a path that belongs to a non trivial cicle, and glue the cicle to your path.
You've obtained a longer path that doesn't intersect the other, against the maximality hypotesis, absurd.
So The theorem is true. 
