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I was reading an article and I have some troubles to understand it. First, the required definition to understand the problem:

Let $mathcal{U}subseteq {Asubseteqomega: |A|=aleph_0 }$. We say that $mathcal{U}$ is an almost disjoint family if for all $A,Binmathcal{U}$ such that $Aneq B$ we have that $|Acap B|<aleph_0$

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

1. 
   $b=d^{*}$.


2. 
   $a=r_alpha$ and $b=d_beta$ with $alphaneqbeta$. Here probably we have two subcases because $alpha<beta$ or $beta<alpha$.


3. 
   $ainomega$ and $b=d_alpha$ for some $alpha<mathfrak{c}$
   


4. 
   $ainomega$ 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.

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 $betaneq alpha$) 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...

Let $f:mathcal Rlongrightarrow Y$ 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 $f={langle r_alpha,d_alpharangle mid alpha<mathfrak{c}}$ is closed in $mathcal Rtimes Y$, which actully follows from this elementary theorem f is Continuous if and only if its Graph is Closed in 𝑋×𝑌 (here we only need the Hausdorffness of $Y$). As $mathcal R$ is closed in $X$, $mathcal R times Y$ is a closed subset of $Xtimes Y$. Hence ${langle r_alpha,d_alpharangle mid alpha<mathfrak{c}}$ is closed in $Xtimes Y$.