Problem: Show that every nontrivial connected graph $G$ has a closed spanning walk that contains every edge of $G$ exactly twice.
My Attempt: I've tried to consider two cases. First $G$ is Eulerian, then we have a Eulerian Circuit $C$, which could simply be traversed again. Next, suppose that $G$ is not Eulerian. In this case, consider the graph $G$ with additional edges $(u,v)$, where $u$ and $v$ are odd vertices. Then we have a graph which has essentially all vertices of even degree and so we obtain a Eulerian Circuit $C_1$. Now we delete all the edges thata re incident with odd vertices from this graph and thus we are left another subgraph $H$ which has all even vertices giving us another circuit $C_2.$ I claim that $C_1$ and $C_2$ cover all edges twice but I can't prove it. Please help.