When the intersection of two sets of second category is of second category? I would like to know if the next result is true.
Proposition: Let $(X, \tau)$ be a compact Hausdorff space and $A$ a dense subset. If $(A, \tau_A)$ is a Baire space and $B$ is of second category in $X$ then $A \cap B$ is of second category in $X$.
I know that proposition is false if $A$ is not a Baire space. For example, $X = [0,1]$, $A = \mathbb{Q} \cap [0,1]$ and $B$ any (nonempty) open subset. But I still can´t prove that.    
 A: Here is a counterexample assuming the continuum hypothesis. (A straightforward generalization works under the weaker hypothesis, that the real line is not the union of fewer than $2^{\aleph_0}$ nowhere dense sets.)
Let $X=[0,1]$ with the usual topology.
Claim: Using transfinite induction we can construct subsets $A,B$ of $X$ such that:
(1) $A\cap B=\emptyset;$
(2) $A$ and $B$ each have uncountable intersection with every nonempty open subset of $X;$
(3) $A$ and $B$ each have countable intersection with every nowhere dense closed subset of $X.$
Proof: List the nowhere dense closed subsets of $X$ in a transfinite sequence $\langle N_\nu:\ \nu\lt\omega_1\rangle.$ List the nonempty open sets in a transfinite sequence $\langle U_\nu:\ \nu\lt\omega_1\rangle$ where each nonempty open $U$ is listed $\aleph_1$ times. At step $\nu\lt\omega_1$ choose
$$a_\nu,b_\nu\in U_\nu\setminus\left(\{a_\mu:\mu\lt\nu\}\cup\{b_\mu:\mu\lt\nu\}\cup\bigcup_{\mu\lt\nu}N_\mu\right),\ a_\nu\ne b_\nu.$$
It follows from (3) that $A$ and $B$ each have countable intersection with every first-category subset of $X.$ Clearly, $B$ is of second category in $X,$ while $A\cap B=\emptyset$ is of first category in $X.$
Claim: $A$ is a Baire space in its relative topology.
Proof: Let $D_1,D_2,\dots,D_n,\dots$ be any sequence of dense open sets in $A.$ For each $n,$ we have $D_n=U_n\cap A$ for some set $U_n$ which is open in $A.$ Moreover, since $D_n$ is dense in $A,$ which is dense in $X,$ the sets $D_n$ and $U_n$ are dense in $X.$ Thus $X\setminus U_n$ is a nowhere dense closed set in $X,$ and $M=X\setminus\bigcap_{n=1}^\infty U_n$ is of first category in $X.$ It follows that $A\setminus\bigcap_{n=1}^\infty U_n=A\cap M$ is countable. Since $A$ is "uncountably dense", it follows that $A\cap\bigcap_{n=1}^\infty U_n$ is dense.
