Let $\mathrm{r}\mathscr{O}$ be the family of open domains (regular open sets) of a topological space $\langle X,\mathscr{O}\rangle$, that is:
$$A\in\mathrm{r}\mathscr{O}\iff A=\mathrm{int(\mathrm{cl(A)})}.$$


A space $\langle X,\mathscr{O}\rangle$ is *normal* iff for any disjoint closed sets $A$ and $B$ there are $O\_1,O\_2\in\mathscr{O}$ such that:
$$
A\subseteq O\_1\quad\text{and}\quad B\subseteq O\_2 \quad\text{and}\quad O\_1\cap O\_2=\emptyset.
$$


Define:
$$\tag{df $\Subset$}
A\Subset B\iff \mathrm{cl}(A)\subseteq B,
$$
and consider the following property:
$$\tag{$\dagger$}
(\forall{A,B\in \mathrm{r}\mathscr{O}})\bigl(A\Subset B\Rightarrow(\exists{C\in \mathrm{r}\mathscr{O}}) (A\Subset C\Subset B)\bigr)
$$
It is easy to prove that if $\langle X,\mathscr{O}\rangle$ is normal, then it satisfies ($\dagger$). What I am interested in is whether the following is true:



> 
> If $\langle X,\mathscr{O}\rangle$ is a Hausdorff space ($T\_2$-space) which satisfies ($\dagger$), then it is normal.
> 


EDIT: Ramiro de la Vega pointed to a very nice counterexample. I have one more question: what if we require that $\langle X,\mathscr{O}\rangle$ is semiregular, that is the regular open sets form a basis for the topology? Thus what I am now asking is whether the following (weaker) statement is true:



> 
> If $\langle X,\mathscr{O}\rangle$ is a semiregular Hausdorff space which satisfies ($\dagger$), then it is normal.
> 


EDIT: The answer to the question above is negative as well. A counterexample is *relatively prime integer topology*, L.A. Steen, J.A. Seebach, Jr. *Counterexamples in topology*, number 60.



Is normality of a Hausdorff space consequence of some property of open domains?
No. 


For example let $X=\{(x,y): x,y \in \mathbb{Q}, y \geq 0 \}$ with the [irrational slope topology](http://brubeck.jdabbs.com/spaces/irrational-slope-topology/). The clousure of any two non-empty open subsets of $X$ have non-empty intersection. This implies that $X$ satisfies ($\dagger$) trivially but it also implies that $X$ is not even $T\_{2\frac{1}{2}}$ (so it is not $T\_4$).


