Separating a set with $\alpha$-property from outside points in a Hausdorff or Urysohn space ‎A subspace ‎‎$‎A‎$‎ of a space X with topology ‎$  ‎\tau‎$‎ ‎has ‎the $\alpha‎$‎-‎property relative to ‎‎$‎X‎$‎ if each
‎$\tau‎$‎-open family which covers ‎‎$‎A‎$‎ has a finite subfamily whose union is ‎$\tau‎$‎‎-dense in $‎A$.‎

I want to show that if $X$ is a Hausdorff space and $A\subset X$ has ‎$ ‎\alpha‎$‎-‎property, then if $x_{0} \in A^{\complement}$, there exist open disjoint subsets $U, V \subseteq X$ such that $x_{0} \in U$ and $A \subseteq V$

My attempt:Since  $X$ is a Urysohn space, for any points $a \in A$ and  $x_{0} \in X$  there exist disjoint open subsets $x_{0} \in U_{a}, a \in V_{a}$ so that $\operatorname{cl}(U_{a}) \cap \operatorname{cl}(V_{a}) = \emptyset $. Consider the collection $ \{ V_{a} | a \in A \}$. This forms an open cover for
$A$ and, since $A$  has ‎$\alpha‎$‎-‎property‎, $\{ V_{a} | a \in A \}$  has a finite subcover, such that $$A \subseteq ‎\bigcup_{i=1}^n‎ \operatorname{‎cl}(‎V‎_{a_i}) ‎=: V‎‎$$ for some $a_1,\ldots,a_n \in A$. Recall that each $V_{a_i} ,(1\leq i \leq n)$ has a corresponding $U_{a_i}$. So consider the intersection of the collection of these $U_{a_i}$: $U:= U_{a_i} \cap \ldots \cap U_{a_n}$. We know that $U$ is open and $x_{0} \in U$, but how to say $V$ is open and $V \cap U = \emptyset$?
If this is true, can Urysohn be replaced by Hausdorff? (In topology, a Urysohn space is a topological space in which any two distinct points can be separated by closed neighbourhoods. Note that a "closed neighbourhood of $x$" is a closed set that contains an open set containing $x$.)
 A: First note that a more common name for a set $A$ with the $\alpha$-property relative $X$ is an "$H$-set in $X$", see this paper by Johannes Vermeer, who quotes an earlier paper by Velicko for this definition.
In a Hausdorff space such an $H$-set is closed, as is easily seen: Fix $p \notin A$ and for each $a \in A$, pick disjoint open neighbourhoods $U_a$ of $a$ and $V_a$ of $p$. As in your attempt, we have finitely many $F \subseteq A$ such that $$A \subseteq \bigcup_{a \in F} \overline{U_a}$$ and then $V=\bigcap_{a \in F} V_a$ is open and misses $A$: if $x \in A \cap V$ then $x \in A$ so $x \in \overline{U_a}$ for some $a \in F$, but $U_a \cap V_a = \emptyset$ implies $\overline{U_a} \cap V_a = \emptyset$ so $x \notin V_a$ and so $x \notin V$ a fortiori, contradiction.
The same argument shows that in your attempt $U \cap V = \emptyset$ too, but your $V$ is a closed set (a finite union of closed sets) so we cannot expect it to be open, and so does not work in general as the required $V$. If we leave out the closures in the union we get an open set, but it need not cover $A$...
So my hunch is that a counterexample to your main question might very well exist in some non-regular, Urysohn $X$ where the counterexample in regularity uses an $H$-set as the closed subset. I'll have a further think.
