Question

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.

Answer 1

No.

For example let $X=\{(x,y): x,y \in \mathbb{Q}, y \geq 0 \}$ with the 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 $T_4$).