Let $(E,\mathcal E)$ be a measurable space, $\mu$ and $\nu$ be $\sigma$-finite measures on $(E,\mathcal E)$, $q\in L^2(\mu\otimes\nu)$ and $$Qg:=\int\nu({\rm d}y)q(\;\cdot\;,y)g(y)\;\;\;\text{for }g\in L^2(\nu).$$
Are we able to show that $Q$ is Hilbert-Schmidt from $L^2(\nu)$ to $L^2(\mu)$?
Let $(g_i)_{i\in I}$ be an orthonormal basis of $L^2(\nu)$. Then \begin{equation}\begin{split}\sum_{i\in I}\left\|Qg_i\right\|_{L^2(\mu)}^2&=\sum_{i\in I}\int\mu({\rm d}x)\left|\langle q(x,\cdot\;),g_i\rangle_{L^2(\nu)}\right|^2\\&\color{red}=\int\mu({\rm d}x)\sum_{i\in I}\left|\langle q(x,\cdot\;),g_i\rangle_{L^2(\nu)}\right|^2\\&=\int\mu({\rm d}x)\left\|q(x,\;\cdot\;)\right\|_{L^2(\nu)}^2=\left\|q\right\|_{L^2(\mu\otimes\nu)}^2<\infty.\end{split}\tag1\end{equation}
The crucial equation is the red one. Which argument do we need?
Remark: Please note that I'm not assuming that $L^2(\mu)$ or $L^2(\nu)$ is separable. Is this a problem? Are we able to conclude that $Q$ is compact? Non-separability shouldn't be a problem, but I'm not sure whether I miss something crucial.