Impossible Covering Properties for Sets of Reals I've been reading more about selection principles (covering properties) recently. Below is terminology. 
Adapting what B. Tsaban said in this article, we consider spaces $X$ which are (homeomorphic to) sets of reals. This will help to filter out problems arising from topologically-pathological examples.
Let $A$ and $B$ be collections of covers of $X$. 


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*$S_1(A, B)$ means for each sequence $\{\mathcal{U}\}_{n \in \mathbb{N}}$ of members of $A$, there are members $U_n \in \mathcal{U}_n$ such that $\{U_n : n \in \mathbb{N}\} \in B$. 


In other words, this builds a new cover for $X$ by picking a single element from each $\mathcal{U}_n$. 
Some covers come into play: 


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*$\mathcal{U}$ is an $\omega$-cover of $X$ if each finite subset of $X$ is contained in some $U \in \mathcal{U}$. 

*$\mathcal{U}$ is a $\gamma$-cover of $X$ if $\mathcal{U}$ is infinite and each $x \in X$ belongs to all but finitely many $U \in \mathcal{U}$. 
By cover, we mean proper covers of $X$ (where $X$ is not a cover of itself). 


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*$O$ is the collection of all open covers of $X$, 

*$\Gamma$ is the collection of all $\gamma$-covers of $X$, 

*$\Omega$ is the collection of all $\omega$-covers of $X$. 


We can also assume that all the covers in these collections are countable. 
In this article by W. Just, A.W. Miller, M. Scheepers, and P.J. Szeptycki, it is mentioned that a couple selection principles of this form cannot hold for certain $X$. 

$S_1(O, \Gamma)$ and $S_1(O, \Omega)$ are impossible for non-trivial $X$. 

I'm having a hard time seeing why these two selection principles are impossible for non-trivial $X$. Is there a property that prevents these two selection principles from happening since we are considering spaces $X$ which are (homeomorphic to) sets of reals?
Any insight would be greatly appreciated. 
 A: It is not very difficult to notice the following.

Observation 1. If $A$ and $B$ are collections of covers of $X$ and there is a cover $\mathcal U\in A$ such that no countable subcover of $\mathcal U$ is in $B$, then $S_1(A,B)$ is false.

Proof. Simply take $\mathcal U_n=\mathcal U$. If $X$ is $S_1(A,B)$ then there is $\{U_n; n\in\mathbb N\}\subseteq \mathcal U$ which belongs to $B$. I.e., we have a countable subcover of $\mathcal U$ which belongs to $B$. $\square$

We may also notice that if $\mathcal U$ has an $\omega$-subcover then $\mathcal U$ itself is an $\omega$-cover. 
It is not clearly stated what it means that $X$ is non-trivial, but we can see that:

Observation 2. If $X$ has an open cover which is not $\omega$-cover, then $X$ does not have property $S_1(\mathcal O,\Omega)$.

Now the only question is whether for non-trivial $X$ there is an open cover which is not $\omega$-cover. Since you mentioned in the question that you are interested only in subspaces of $\mathbb R$ (and their subspaces), this should be sufficient.

Observation 3. If $X$ is a regular Hausdorff space which has at least two points, then there is an open cover of $X$ which is not $\omega$-cover.

Proof. Simply take some two points $a\ne b$. Then we have an open set $V$ such that $a\in V$ and $b\notin V$.
From regularity we also have another open set $U$ such that $a\in U$ and $\overline U\subseteq V$.
Then ve have $\mathcal U=\{V, X\setminus \overline U\}$ is an open cover of $X$. But there is no set in $\mathcal U$ which contains both $a$ and $b$. $\square$

Since every $\gamma$-cover is also $\omega$-cover, we have $S_1(\mathcal O,\Gamma) \Rightarrow S_1(\mathcal O,\Omega)$. So if $S_1(\mathcal O,\Omega)$ is impossible, so is $S_1(\mathcal O,\Gamma)$. 
