An implication for covering properties on the reals To help get a better idea of how covering properties (Selection Principles) work, I started to prove known results that are given without proof. 
Some background information:


*

*$X$ is a set of reals and both $\mathcal{A}$ and $\mathcal{B}$ are collections of covers of $X$. In the context of the article where the problem is mentioned,  $\mathcal{A}$ and $\mathcal{B}$ can be open covers, open large covers,
$\omega$-covers, or $\gamma$-covers of $X$.

*$S_{fin} (\mathcal{A}, \mathcal{B})$ means for each sequence $\{U_n\}_{n \in \mathbb{N}}$ of members of $\mathcal{A}$, there exists finite subsets $F_n$ of $U_n$, $n \in \mathbb{N}$, such that $\bigcup_{n \in \mathbb{N}} F_n \in \mathcal{B}$. 

*$U_{fin} (\mathcal{A}, \mathcal{B})$ means for each sequence $\{U_n\}_{n \in \mathbb{N}}$ of members of $\mathcal{A}$  which do not contain a finite sub-cover, there exists finite subsets $F_n$ of $U_n$, $n \in \mathbb{N}$, such that $\{\bigcup F_n : n \in \mathbb{N}\} \in \mathcal{B}$
I am trying to show

$$S_{fin} (\mathcal{A}, \mathcal{B}) \Rightarrow U_{fin} (\mathcal{A}, \mathcal{B}).$$

First, I let $\{U_n\}_{n \in \mathbb{N}}$ be a sequence of members of $\mathcal{A}$, where each $U_n$ does not contain a finite sub-cover. Since $S_{fin} (\mathcal{A}, \mathcal{B})$ holds, I'm able to find finite $F_n \subset U_n$ such that  $\bigcup_{n \in \mathbb{N}} F_n \in \mathcal{B}$. 
I keep getting stuck from here. I'm not seeing how each $U_n$ not having a finite sub-cover will lead us to $U_{fin}$. I'm also having a hard time visualizing what   $\bigcup_{n \in \mathbb{N}} F_n$ and $\{\bigcup F_n : n \in \mathbb{N}\}$ mean in general. 
Anything to help get me in the right direction would be really appreciated.
 A: As to the last remarks: $U_n$ is itself an open cover of $X$ (of a certain type: it belongs to $\mathcal{A}$ and could have the additional assumption of no finite subcovers) and $F_n$ is a finite subfamily of it. So $F_n$ consists of finitely many open subsets of $X$. Then $\cup_n F_n$ is just the family of open subsets of $X$ that we get when we collect all those sets together. The claim then is twofold: this family is a cover of $X$ again (which it need not be if we choose unwisely) and it belong to the target family $\mathcal{B}$ as well.
$\{\cup_n F_n: n \in \omega \}$ is a different animal: here we take the union of the $F_n$, which is some (open) subset of $X$, so we get one open unioned set per cover $U_n$, which is not a member of $U_n$ necessarily (and which will not be $X$ if the $U_n$ have no finite subcovers), but such that the collection of these new (open, finite union) sets is again a cover of $X$ and belongs to $\mathcal{B}$.
In most papers on the properties (and the one you link to in particular) people consider 4 main classes of open covers: 


*

*$\mathcal{O}$, the set of all open covers of $X$.

*$\Gamma$, the set of all $\gamma$-covers of $X$: these are all infinite open covers $\mathcal{U}$ of $X$ such that every $x$ in $X$ is in almost all members, or more formally, $\mathcal{U}$ is an open cover of $X$, $|\mathcal{U}| \ge \omega$ and $\forall x \in X: \left|\{U \in \mathcal{U}: x \notin U \}\right| < \omega$.

*$\Omega$, the set of all $\omega$-covers of $X$. $\mathcal{U}$ is an $\omega$-cover of $X$ iff it is an open cover that has no finite subcover and for all finite subsets $F$ of $X$, there is an $U \in \mathcal{U}$ such that $F \subseteq U$.

*$\Lambda$, the set all large covers of $X$, which means that every point of $X$ is in infinitely many members of the cover.


Then $\Gamma \subseteq \Omega \subseteq \Lambda \subseteq \mathcal{O}$. The last one holds because we only consider open covers of $X$ (most papers consider Lindelöf spaces $X$ and assume all these covers are countable ones, often also never containing $X$ ("non-trivial covers") and even not containing finite subcovers). 
A $\gamma$-cover is an $\omega$-cover, because if $F$ is a finite subset of a $\gamma$-cover, finitely many $U$ do not contain all of the members of $F$ (a finite union of finitely many sets missing single members is still finite) and the cover is infinite, so there are sets left in the cover, all of whom must contain all of $F$. 
An $\omega$-cover is large, because if some $x$ were only in the sets $U_1,\ldots,U_n$, these sets would not form a finite subcover, and then for any $y \in X \setminus \cup_{i=1}^n U_i$, the finite set $\{x,y\}$ cannot be contained in one member of the cover (such a member must be one of the $U_i$ but none of these contain $y$).
It's also clear that for any class of covers $\mathcal{B}$ among those 4 then 
$$ S_{fin}(\mathcal{O},\mathcal{B}) \rightarrow S_{fin}(\Lambda,\mathcal{B}) \rightarrow S_{fin}(\Omega,\mathcal{B}) \rightarrow S_{fin}(\Gamma,\mathcal{B}) $$
because the sets of the covers we need to consider gets smaller.
Similarly, the property gets weaker as the target cover class gets larger, so for all classes $\mathcal{A}$ among these four we have:
$$ S_{fin}(\mathcal{A}, \Gamma) \rightarrow S_{fin}(\mathcal{A}, \Omega) \rightarrow S_{fin}(\mathcal{A}, \Lambda) \rightarrow S_{fin}(\mathcal{A}, \mathcal{O}) $$
and note that the same holds for the corresponding $U_{fin}(\mathcal{A},\mathcal{B})$ properties.
Also, for countable covers (as is usual) we can go back too, for the $U_{fin}$ poperties: 
$$U_{fin}(\Gamma,\mathcal{B}) \rightarrow U_{fin}(\mathcal{O}, \mathcal{B})$$ 
This holds because if we have any countable open cover $U_n, n \in \omega$, then the cover defined by $V_n = \cup_{i=0}^n U_i$ is a $\gamma$-cover (for every $x$ there is a first $n$ such that $x \in U_n$, and then $x$ is in all the $V_k$ for $k \ge n$). And a union of finitely many sets from the $V_n$-cover is a union of finitely many sets of the original $U_n$-cover as well. So we can transform a sequence of countable open covers to their "$\gamma$-fications", apply the principle to those, and choose the corresponding sets to get a cover in $\mathcal{B}$.
What do we need to get 
$$S_{fin}(\Gamma, \mathcal{A}) \rightarrow U_{fin}(\Gamma, \mathcal{A})$$ 
for different classes of covers $\mathcal{A}$?
So we start with a sequence of $\gamma$-covers $\mathcal{U}_n$, all without finite subcovers, so from the $S_{fin}$ we get finite subsets $\mathcal{F}_n \subseteq \mathcal{U}_n$ such that $\cup_n \mathcal{F}_n$ is in $\mathcal{B}$.
If $\mathcal{B} = \mathcal{O}$, then the same $\mathcal{F}_n$ are as required for the $S_{fin}$ property, because then all we want is to be an open cover, and the finite unions (as subsets of $X$, recall) $\cup \mathcal{F_n}$ also form an open cover (open sets are preserved by unions and $\cup_n (\cup \mathcal{F}_n) = \cup \cup_n \mathcal{F}_n = X$.
If $\mathcal{B} = \Omega$ then again the same $\mathcal{F}_n$ work, because if $F$ is any finite subset of $X$, then some member of $\cup_n \mathcal{F}_n$ contains $F$, and this member is thus a member $U$ of $\mathcal{F}_n$ in particular, and so $F \subseteq \cup \mathcal{F}_n$, which shows that the set of unions of $\mathcal{F}_n$'s is an $\omega$-cover as well. And a finite subcover of the "cover of unions" would give a finite subcover in the union of $\mathcal{F}_n$ cover as well, so cannot exist. 
If $\mathcal{B} = \Gamma$, then the same $\mathcal{F}_n$ work again. If $x \in X$, then as $\cup_n \mathcal{F}_n$ is a $\gamma$-cover, $x$ is only missed by finitely many of the members of all the $\mathcal{F}_n$. So it is missed by possibly even fewer of the $\cup \mathcal{F}_n$. So the latter is also a $\gamma$-cover.
The case $\mathcal{B} = \Lambda$ is a bit trickier. Taking the finite unions $\cup \mathcal{F}_n$ might leave us with fewer sets than expected (as unions need not be distinct). The $\mathcal{F}_n$ (from the $S_{fin}$) give us a large cover, so every $x$ is in infinitely many sets from $\cup_n \mathcal{F}_n$. Then we define finite $\mathcal{G}_n$, increasing in $n$, with $\mathcal{F}_n \subseteq \mathcal{G}_n \subseteq \mathcal{U}_n$, such that for all $n$ and for all $m < n$: $\cup \mathcal{G}_n  \neq \cup \mathcal{G}_m$: define $\mathcal{G}_0 = \mathcal{F}_0$ and given $\mathcal{G_n}$ we note that $\cup (\mathcal{G}_n \cup \mathcal{F}_{n+1}) \neq X$ (as the $\mathcal{U}_n$ have no finite subcover), so pick $y$ not in the union, which is covered by some $O \in \mathcal{U}_n$, and define $\mathcal{G}_{n+1} = \mathcal{G} \cup \mathcal{F}_{n+1} \cup \{O\}$. 
This ensures that the sets $\cup_n G_n$ form a large cover (as they're all distinct).  
So for all 4 classes $\mathcal{A}, \mathcal{B}$ we have 
$$S_{fin}(\mathcal{A},\mathcal{B}) \rightarrow S_{fin}(\Gamma, \mathcal{B}) \rightarrow U_{fin}(\Gamma, \mathcal{B}) \rightarrow U_{fin}(\mathcal{A}, \mathcal{B})$$
