Let $\text{BP}(X)$ denote $\sigma$-algebra of subsets of $X$ with the Baire Property BP and $\text{MGR}(X)$ denote the $\sigma$-ideal of meager sets in $X$.
Assume $X$ is second countable Baire space.
Question: There is no uncountable subset $\mathcal{A} \subseteq \text{BP}(X)$ such that $A \notin \text{MGR}(X)$ for any $A \in \mathcal{A}$ and $A\cap B \in \text{MGR}(X)$ for anytwo distinct $A,B \in \mathcal{A}$.
Suppose I have uncountably many non-meager sets $\mathcal{A}=\{A_\eta\}_{\eta\in\omega_1}$ with the Baire property. Let $\mathcal{U}$ be a countable base for $X$.
Let $\mathcal{U}$ be a countable base for the space $X$. Given any non-meager $A\in\mathcal{A}$, there is some open set $V\in\mathcal{U}$ and some meager set $D$ such that $V\Delta D\subseteq A$: $A$ has the property of Baire, so differs from some open set by a meager set, and I'm just restricting my attention to a small piece of that open set. Without loss of generality, I'll assume each $A\in\mathcal{A}$ actually equals $V\Delta D$ for some $V\in\mathcal{U}$ and some meager $D$.
Now since $\mathcal{U}$ is countable there is some nonempty open set $O\in \mathcal{U}$ and uncountable $\mathcal{B}\subseteq\mathcal{A}$ such that for each $B\in\mathcal{B}$, $B\Delta O$ is meager. In particular, since $\mathcal{B}$ is uncountable, it has at least two elements. Pick some $B_0, B_1$ in $\mathcal{B}$. We have $B_0\cap B_1\supseteq O-(M_0\cup M_1)$ for meager sets $M_0, M_1$. But since $X$ has the Baire property, this intersection is non-meager.
