Let $G$ be a connected graph, show that $G$ has $2k>0$ vertices of odd degree if and only if there exists a partition of the edges of $G$ into $k$ open Eulerian walks.
Attempt:
$\Rightarrow )$ Let $y_1, y_2, \ldots y_k$ and $z_1, z_2, \ldots , z_k$ be the odd vertices of $G$. We construct a new graph $H$ from $G$ by adding $k$ new vertices $x_1, x_2, \ldots , x_k$ to $G$ and joining $x_i$ to $y_i$ and $z_i$ for $i = 1, 2, \ldots , k$. Thus, $H$ is Eulerian and therefore contains an Eulerian circuit $C$. Since $y_ix_i$ and $x_iz_i$ are consecutive on $C$ for $i = 1, 2, \ldots ,k$, deleting the $k$ vertices $x_i$ from $H$ results in $k$ edge-disjoint trails in $G$ connecting odd vertices such that every edge of $G$ lies on one of these trails, i.e., they are an open Eulerian trail.
$\Leftarrow )$ Suppose that $G$ has a partition in edge-disjoint trails. Let $P_1, P_2, \ldots P_k$ be the walks contained in G such that $\cup _{i=1}^{l}E(P_i)=E(G)$. Since $G$ is connected, the union of the open trails gives a closed trail $P$, since $E(P_i)\cap E(P_j)=\emptyset$ for all $i\neq j$. Thus $P$ is a closed Eulerian trail.
Is what I did okay?