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 $\mathscr{U}$.