Is there a countable pseudocharacter Hausdorff space X with the property A? Let X be a Hausdorff space and define the Property A as follows: 

If $\mathscr{U}$ is a collection of open sets of $X$ that witnesses Hausdorff property of $X$ (= $\forall x,y \in X$, there exist two disjoint open sets $U_1$ and $U_2$ $\in \mathscr{U}$ s.t. $x \in U_1$ and $y \in U_2$), then there is a point $x\in X$ such that $|\{U \in \mathscr{U}: x \in U\}|>\omega$. 

Is there a countable pseudocharacter Hausdorff space $X$ with the property A? The same question is also asked here.
 A: Let $X = \beta\omega$ with the following topology: points of $\omega$ are isolated, and each $p\in\beta\omega\setminus\omega$ has $\{\{p\}\cup U:U\in p\}$ as a local base. (I think of the points of $\beta\omega\setminus\omega$ as free ultrafilters on $\omega$.) 
This topology refines the Čech-Stone topology on $\beta\omega$, so it’s certainly Hausdorff, and since it has a base of countable sets, it must have countable pseudocharacter as well. But $\beta\omega\setminus\omega$ is uncountable (in fact of cardinality $2^{2^\omega}$), so any family $\mathscr{U}$ that witnesses the Hausdorffness of $X$ must be uncountable. 
Added: That actually doesn’t follow without further argument. However, $|\beta\omega|>2^\omega$, so if $\mathscr{U}$ is a countable family of subsets of $\beta\omega$, there are distinct $p,q\in\beta\omega$ such that $$\{U\in\mathscr{U}:p\in U\}=\{U\in\mathscr{U}:q\in U\},$$ and $\mathscr{U}$ cannot then witness the Hausdorffness of $X$.
Each member of $\mathscr{U}$ has non-empty intersection with $\omega$, so some $n\in\omega$ must belong to uncountably many members of $\mathscr{U}$.
