Product of Mrówka space and one point compactification discrete space. I was reading an article and I have some troubles to understand it. First, the required definition to understand the problem:

Let $\mathcal{U}\subseteq \{A\subseteq\omega: |A|=\aleph_0 \}$. We say that $\mathcal{U}$ is an almost disjoint family if for all $A,B\in\mathcal{U}$ such that $A\neq B$ we have that $|A\cap B|<\aleph_0$

The proof that I was reading is the next:

The key part of the proof is the fact that $A$ is a closed subset of $X\times Y$. But I can't see that $A$ is closed only by the construction of the topology of $X\times Y$. In fact, I think that we need a lot of cases to prove that fact because if we take $(a,b)\in (X\times Y)\setminus A$ then


*

*$b=d^{*}$. 

*$a=r_\alpha$ and $b=d_\beta$ with $\alpha\neq\beta$. Here probably we have two subcases because $\alpha<\beta$ or $\beta<\alpha$.

*$a\in\omega$ and $b=d_\alpha$ for some $\alpha<\mathfrak{c}$

*$a\in\omega$ and $b=d^*$.


Are they all cases? Or am I forgetting some? I don't know if my thoughts are correct. Can you help me to complete the proof? I really appreciate any help you can provide me. 
 A: Suppose that $(x,y) \notin A$. If $y \neq d^\ast$, then $y=d_\alpha$ for some $\alpha < \mathfrak{c}$ while then $x \neq r_\alpha$. But then taking the neighbourhoods $U_x=\{r_\beta\} \cup r_\beta$ (if $x=r_\beta$ for some $\beta\neq \alpha$) or $U_x =\{x\}$ (if $x \in \omega$) and $V_y=\{d_\alpha\}$ we have that $U_x \times V_y$ also misses $A$, as $V_y$ only contains $y$ so the only way it could intersect $A$ is when $r_\alpha \in U_x$ which is clearly not the case by construction. 
So the case that $y \neq d^\ast$ has been covered. So suppose $y=d^\ast$ and we need a neighbourhood of $(x,y)$ that misses $A$. If $x \in \omega$ take $\{x\} \times Y$ which clearly works as $A$ has no points with first coordinate in $\omega$, and if $x=r_\beta$ for some $\beta$, then it's easy to see that $(\{r_\beta\}\cup r_\beta) \times (Y \setminus \{ d_\beta \})$ is basic open and misses $A$ (as the neighbourhood of $r_\beta$ contains no other $r_\alpha$ by definition, just $r_\beta$ and some isolated points in $\omega$).
Just a word of warning:
Lemma 2.1 is false, and $X$ itself is a counterexample: $\mathcal{R}$ is closed and uncountable and discrete so not weakly Lindelöf. So not every closed subset of $X$ is weakly Lindelöf, so $X$ is separable ($\omega$ is dense) but not quasi-Lindelöf.
True is: a separable space is weakly Lindelöf. This is rather trivial to prove. The reference is not from a "proper" journal, so be warned...
Theorem 3.37 in the referenced paper [5] is also false (ccc implies quasi-Lindelöf) by the same counterexample. Don't believe everything in every random journal... I think this is the source of this paper's lemma 2.1 as I did not find a theorem on separable spaces (but of course separable implies ccc).  
Also: 3.36 in [5] (every weakly Lindelöf normal space is quasi-Lindelöf) has a ZFC counterexample: $C_p(X)$ where $X$ is the one-point Lindelöfication of an uncountable discrete space (which can be seen to be the same as the $\Sigma$-product of uncountably copies of $\mathbb{R}$). Then $C_p(X)$ is (collectionwise) normal and is ccc; (it even has a dense Lindelöf subspace,) hence is weakly Lindelöf, but has a closed subspace homeomorphic to $\omega_1$ and hence is not quasi-Lindelöf (the cover by intial segments witnesses that $\omega_1$ is not weakly Lindelöf). For more  details see this blog post, an excellent source for examples as these...
A: Let $f:\mathcal R\longrightarrow Y$ be $f(r_\alpha)=d_\alpha$. $f$ is clearly continuous because $\mathcal R$, as a subspace of $X$, is discrete. Now, we claim that the graph of $f=\{\langle r_\alpha,d_\alpha\rangle \mid \alpha<\mathfrak{c}\}$ is closed in $\mathcal R\times Y$, which actully follows from this elementary theorem f is Continuous if and only if its Graph is Closed in × (here we only need the Hausdorffness of $Y$). As $\mathcal R$ is closed in $X$, $\mathcal R \times Y$ is a closed subset of $X\times Y$. Hence $\{\langle r_\alpha,d_\alpha\rangle \mid \alpha<\mathfrak{c}\}$ is closed in $X\times Y$.
