Is the classical Mrowka space $\Delta$-normal? Definition: A space $X$ is $\Delta$-normal if for every $A \subset X^2 \setminus \Delta_X$ closed in $X^2$ there exist disjoint open $U$ and $V$ in $X^2$ such that $A \subset U$ and $\Delta_X \subset V$, where $\Delta_X$ is the diagonal of $X$, i.e., $\{(x,x): x\in X\}$.
Is Mrowka space $\Delta$-normal?
Thanks for your help.
 A: Few days ago I told about your question to Serhii Bardyla, and a few minutes ago we showed that the Mrówka space (with the maximal almost disjoint family $\mathcal A$) is not $\Delta$-normal. Put 
$$F=(\mathcal A\times  \mathcal A)\setminus\Delta_X=(\mathcal A\times  \mathcal A)\setminus\bigcup_{A\in\mathcal A}(\{A\}\cup\omega)\times (\{A\}\cup\omega).$$
Thus $F$ is a closed subset of $X\times X$ disjoint from $\Delta_X$.
Let $U$ and $V$ be open in $X\times X$ such that $F\subset U$ and $\Delta_X \subset V$. Let $\mathcal A’=\{A_n\}$ be any sequence of distinct elements of $X$. Since $V$ is a neighborhood of $\Delta_X$, for each $n\in\omega$ there exists a number $N_n\in\omega$ such that $V_n=(\{A_n\}\cup (A_n\setminus N_n))\times (\{A_n\}\cup (A_n\setminus N_n))\subset V$. This allows us to choose a sequence $\{k_n\}$ such that $k_{n}\in A_n\setminus (N_n\cup\bigcup_{m<n} A_m)$ for each $n$. Since the family $\mathcal A$ is maximal almost disjoint, there exists $B\in\mathcal A$ such that a set $S=\{n\in\omega: k_n\in B\}$ is infinite. Since $k_n\not\in A_m$ for each $n>m$, $B\not\in\mathcal A’$. Clearly, the sequence $\{k_n:n\in S\}$ converges to $B$. Since the family $\mathcal A$ is maximal almost disjoint, for each $n\in S$ there exists $l_n\in A_n\setminus (N_n\cup B\cup\{l_m: 0\le m<n\})$. Since the family $\mathcal A$ is maximal almost disjoint, there exists a set $C\in\mathcal A$ such that a set $\{n\in S: l_n\in C\}$ is infinite. The choice of the sequence $\{l_n\}$ assures that $C\ne B$. By the construction, 
$$(B,C)\in\overline{\bigcup_{n\in S} V_n}\subset \overline{V}.$$
That is $(B,C)\in F\cap \overline{V}$. Since $F\subset U$, $U$ is a neighborhood of $(B,C)$, so $U\cap V\ne\varnothing$.
