Non meager rectangle Suppose $G \subseteq \mathbb{R}^2$ is dense $G_\delta$. Must there (in ZFC) exist non meager sets of reals $A, B$ such that $A \times B \subseteq G$?
 A: I pointed this question to Taras Banakh,
who pointed to me his joint paper with Lyubomyr Zdomskyy “Non-meager free sets for meager relations on Polish spaces” which contains an answer. 
Abstract. We prove that for each meager relation $E\subset X\times X$ on a Polish space $X$ there is a nowhere meager subspace $F\subset X$ which is $E$-free in the sense that $(x,y)\not\in E$ for any distinct points $x,y\in F$.
From myself I add a simple negative answer for descriptively good $A$ and $B$, that is when both $A$ and $B$ have the Baire Property. I recall, that a subset $B$ of a topological space $X$ has the Baire Property (BP) in $X$ if $B$ contains a $G_\delta$-subset $C$ of $X$ such that $B\setminus C$ is meager in $X$. By [Kech, 8.22] each Borel subset of a space $X$ has the Baire Property in $X$. Put $G=\mathbb{R}^2\setminus \{(x,y):x-y\in\Bbb Q\}.$ Assume that both $A$ and $B$ are non meager subsets of the space $\Bbb R$ with the Baire Property and $A\times B\subset G$. By [Kech, Prop. 8.22] both sets $A+\Bbb Q$ and $B+\Bbb Q$ have the Baire Property. By [Kech, Th. 8.46] both sets $A+\Bbb Q$ and $B+\Bbb Q$ are comeger. So $A+Q\cap B+Q\ne\varnothing$. That is, there exist points $a\in A$, $b\in B$, $q, q’\in\Bbb Q$ such that $a+q=b+q’$. Then $(a,b)\in A\times B\setminus G$, a contradiction. 
References
[Kech] A. Kechris. Classical Descriptive Set Theory, – Springer, 1995.


