Mrówka ($\Psi$) Space $\newcommand{\cl}{\operatorname{cl}}$Suppose we have $\mathbb{N}$ and a collection $M$ of infinite subsets of $\mathbb{N}$ that is almost disjoint: for every $A,B\in M,A\ne B$ we have that $A∩B$ is finite. Furthermore, consider $M$ to be a maximal almost disjoint collection of infinite subsets of $\mathbb{N}$.
Define a topology on a space $X=\mathbb{N}∪\{x_A:A\in M\}$ by defining open sets as such: for each $n\in\mathbb{N}$, $\{n\}$ is open in $X$ and for every $x_A$ (for a fixed $A\in M$) , $B(A,F)=\{x_A\}\cup(A\setminus F)$, where $F\subseteq A$ is finite, is a basic open set of $X$.
If $U$ is an open set of $X$ and $\{x_A:A\in M\}\subseteq U$, prove that $X\setminus \cl_XU$ is finite.
Suppose $X\setminus \cl_XU$ is infinite. Since we know that $X\setminus \cl_XU$ is a subset of $\mathbb{N}$, my approach is to "adjoint" $X\setminus \cl_XU$ to the family $\{x_A:A\in M\}$ and then get a contradiction that $M$ is maximal. But I am stuck in proving $X\setminus \cl_XU \cap A$ to be finite. Can anyone give me some hints?
 A: $\newcommand{\cl}{\operatorname{cl}}$I think that you mean that if $U$ is open in $X$ and $\cl_XU\supseteq\{x_A:A\in M\}$, then $X\setminus\operatorname{cl}_XU$ is finite: $M$ itself isn’t a subset of $X$.
It isn’t necessarily true. Let $E=\{2n:n\in\Bbb N\}$, let $O=\Bbb N\setminus E$, and let $\mathscr{M}_E$ be an infinite maximal almost disjoint family of infinite subsets of $E$. Let $\mathscr{M}_O=\{A+1:A\in\mathscr{M}_E\}$, where $A+1=\{n+1:n\in A\}$; clearly $\mathscr{M}_O$ is a maximal almost disjoint family in $O$. 
For $A\in\mathscr{M}_E$ let $\hat A=A\cup(A+1)$, and let $\mathscr{M}=\{\hat A:A\in\mathscr{M}_E\}$. If $A,B\in\mathscr{M}_E$ with $A\ne B$, then 
$$\begin{align*}
\hat A\cap\hat B&=(A\cap B)\cup\big((A+1)\cap(B+1)\big)\\
&=(A\cap B)\cup\big((A\cap B)+1\big)\;,
\end{align*}$$
which is finite, since $A\cap B$ is finite. Thus, $\mathscr{M}$ is almost disjoint. If $S\subseteq\Bbb N$ is infinite, then at least one of $S\cap E$ and $S\cap O$ is infinite, so there is an $A\in\mathscr{M}_E$ such that either $S\cap A$ is infinite or $S\cap(A+1)$ is infinite. In either case there is an $A\in\mathscr{M}$ such that $S\cap A$ is infinite, so $\mathscr{M}$ is maximal almost disjoint.
Now use $\mathscr{M}$ to form the $\Psi$-space $X$, and let $U=E$; $U$ is certainly open. Let $A\in\mathscr{M}$; then $A\cap U\in\mathscr{M}_E$, so $A\cap U$ is infinite, and $x_A\in\cl_XU$. Thus, $\cl_XU=U\cup\{x_A:A\in\mathscr{M}\}$, and $X\setminus(\cl_XU)=O$, which is infinite.
