Is there a countable pseudocharacter Hausdorff  space，such that...? Let X be a Hausdorff space and Difine the Property A as following: 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 opensets $U_1$ and $U_2$ $\in \mathscr{U}$, st $x \in U_1$ 
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?
 A: Hodel defined the Hausdorff pseudo-character $H\psi(X)$ as the smallest infinite cardinal so that each point lies in at most $H\psi(X)$ elements of an open family $\mathcal{U'}$ that witnesses the Hausdorff property of $X$. Any $X$ such that $\omega = \psi(X) < H\psi(X)$ would seem to answer your question. I don't know of any specific example, but I guess there must be one. Why define $H\psi$ if it's really the same as $\psi$?
A: Just as I suspected -- $\omega_1$ answers the question as it now stands. It is obviously first countable, hence of countable pseudocharacter, but has no point-countable $T_2$-separating open cover.
Let $\mathcal U$ be a point-countable open cover of $\omega_1$. Let $N$ be the collection of all points which are contained in a nonstationary member of $\mathcal U$. I claim $N$ is nonstationary; if not, then the pressing-down lemma would give us a point that is in countably many nonstationary open sets that cover a stationary set, which is impossible.
But now, a stationary open set is actually co-countable, and so every point in the complement of $N$ is contained only in co-countable members of $\mathcal U$, and no two of these can be in disjoint members of $\mathcal U$.
I suspect there is a general theorem involving spaces with point-countable $T_2$-separating open covers that eliminates $\omega_1$ and many other spaces; but I'd have to review the literature on this. 
