Prove a graph Containing $2k$ odd vertices contains $k$ distinct trails
Solution 1:
I see my problem, I'm thinking in terms of simple graphs, but this theorem is thinking in terms of multigraphs. If you add an edge between 3 and 1 and then between 1 and 1 then all vertices become even and there is an Eulerian cycle in the graph. Take away the new edges and you get back a graph in which each edge between odd degree vertices becomes it own trail.
Solution 2:
Here's an alternative proof. Take an odd vertex, and construct a trail from that to another odd vertex. Remove the edges of that trail; the resulting graph has $2k-2$ odd vertices. Repeat.