Okay, so the idea is there, but writing it out is causing trouble. When that happens, it's usually a good idea to write down things carefully.
Let $v$ be a vertex in $G$, and let $e_1,\ldots,e_n$ be all the edges of $G$ that are incident in $v$. Let $m_i$ be the number of times that $e_i$ is used in the walk $W$ (counting multiplicity).
Because the walk is closed, the number of edges in $W$ (counting multiplicity) that are incident in $v$ is even. That means that $m_1+\cdots+m_n \equiv 0 \pmod{2}.$
Reordering the edges if necessary, assume that $m_1,\ldots,m_k$ are odd, $0\leq k\leq n$, and $m_{k+1},\ldots,m_n$ are even. That means that of all $n$ edges incident in $v$, only $e_1,\ldots,e_k$ lie in $H$. You need to show that $k$ is even. Now note that $m_i\equiv 1\pmod{2}$ if $1\leq i\leq k$, and $m_i\equiv 0\pmod{2}$ if $k\lt i\leq n$.