Chain rule for KL divergence
I don't think the given inequality is true (I am not sure about the reverse inequality). Recall that $$
D_{KL}(q(x,y)\| p(x,y))=\sum_{x\in \mathcal{X}}\sum_{y\in \mathcal{Y}} q(x,y)\log\bigg(\frac{q(x,y)}{p(x,y)}\bigg).
$$
Consider non-identical joint distributions $p$ and $q$ over $\mathcal{X}\times\mathcal{X}$ (so $\mathcal{Y}=\mathcal{X}$), with $p(x,y)=q(x,y)=0$ for $y\neq x$ (so the supports of both distributions are on the diagonals). Then 
\begin{align}
D_{KL}(q(x,y)\|p(x,y))&amp;=\sum_{x_1\in \mathcal{X}}\sum_{x_2\in \mathcal{X}}q(x_1,x_2)\log\bigg(\frac{q(x_1,x_2)}{p(x_1,x_2)}\bigg)\\
&amp;=\sum_{x\in \mathcal{X}} q(x)\log\bigg(\frac{q(x)}{p(x)}\bigg)\\
&amp;=D_{KL}(q(x)\|p(x))=D_{KL}(q(y)\|p(y)).
\end{align}
In general, these are not all zero or infinite, so the stated inequality will not hold.

However, I think one can say that 
$$
D_{KL}(q(x,y)\|p(x,y))\geq \frac{D_{KL}(q(x)\|p(x))+D_{KL}(q(y)\|p(y))}{2}.
$$
This should follow from the Chain Rule:
\begin{align}
D_{KL}(q(x,y)\|p(x,y))&amp;=D_{KL}(q(x)\|p(x))+D_{KL}(q(y\vert x)\|p(y\vert x))\\
&amp;=D_{KL}(q(y)\|p(y))+D_{KL}(q(x\vert y)\|p(x\vert y)),
\end{align}
summed twice and then using the nonnegativity of divergence.