$\omega$-covers and $S_1(\Omega,\Gamma)$ property I am reading this article1 and there is some proof in there (top of page 156) which is not clear to me. The definitions are:

1. Property ($\gamma$): If $\mathcal U$ is an $\omega$-cover of $X$, then there is a sequence $\{ U_n : \forall n \in \omega, U_n \in \mathcal U \}$ such that $\underline {Lim}U_n = X $
2. Property ($\gamma'$): If $\mathcal U_n$ is a sequence of $\omega$-covers of $X$, then there is a sequence $\{ U_n : \forall n \in \omega, U_n \in \mathcal U_n \}$ such that $\underline {Lim}U_n = X $
3. If $\langle A_n : n \in \omega  \rangle$ is a sequence of subsets of a set $X$, 
  $$ \underline{Lim} A_n = \{ x \in X : \exists n_0 \in \omega \forall n \geq n_0, x \in A_n \} $$
  If $\mathcal A$ is a family of subsets of a set $X$, then, $L(\mathcal A)$ denotes the smallest family of subsets of $X$ containing $\mathcal A$ and closed under $\underline{Lim}$.

In the paper it is clamed that Property $\gamma$ implies property $\gamma'$.
The proof is as follows:
Let $\mathcal I_n$ be a sequence of $\omega$-covers of $X$. Take an infinite sequence 
$\{ x_n \}$ in $X$ (we are assuming $X$ to be infinite). Define $\mathcal U_n = \{ G \setminus \{ x_n \} : G \in \mathcal I_n\}$ and $\mathcal U = \bigcup \{ \mathcal U_n : n \in \omega \}$. Evidently, $\mathcal U$ is an $\omega$-cover, by property $\gamma$ there is a countable set $U_{n_k}$ such that, $\underline{Lim} U_{n_k} = X$. the set of indexes $n_k$, can not be bounded by any $m \in \omega$ because if it was, then no set from the cover will contain $\{ x_1,...,x_m \}$.
The thing which is not clear to me is, how can we be sure that the new sequence $U_{n_k}$ contains no more then one set from each cover $\mathcal I_n$? 
Any help?
Thank you!
1J. Gerlits, Zs. Nagy: Some properties of $C(X)$, I. Topology and its Applications, Volume 14, Issue 2, 1982, Pages 151-161. DOI: 10.1016/0166-8641%2882%2990065-7
 A: Let's call a countably infinite cover $\{U_n:n<\omega\}$ a gamma cover if $\operatorname{Lim} U_n=X$, put another way, each $x\in X$ belongs to all but finitely many $U_n$. All gamma covers are omega covers, since each member of a finite set belongs to all but finitely many sets out of the infinite collection.
Let's also note that any infinite subset of a gamma cover is also a gamma cover, and if each set in a gamma cover is increased in size we still have a gamma cover.
Now, we should first refine our sequence $\mathcal I_n$ of omega covers by letting $\mathcal I_0'=\mathcal I_0$ and $\mathcal I_{n+1}'=\{U\cap V:U\in \mathcal I_{n+1},V\in \mathcal I_n'\}$, so $\mathcal I_{n+1}'$ also refines $\mathcal I_n'$. As you say, we then let $\mathcal U_n=\{I\setminus\{x_n\}:I\in\mathcal I_n'\}$. So when we apply property $\gamma$ to the resulting omega cover $\mathcal U=\bigcup\{\mathcal U_n:n<\omega\}$ we get a gamma cover $\mathcal U'=\bigcup\{\mathcal U_n':n<\omega\}$ where $\mathcal U_n'\subseteq \mathcal U_n$.
Now if $|\mathcal U_n'|=1$ for all $n$, we'd basically be done. But there's no reason this must be true. So instead, we have to note that since $\mathcal U'$ is an omega cover, for each $n<\omega$ we have a set containing $\{x_m:m\leq n\}$, which cannot belong to $\mathcal U_m$ for any $m\leq n$. Thus we have a cofinal sequence $U_{k_n}\in\mathcal U_{k_n}'$ for $n\leq k_n<\omega$, and $\{U_{k_n}:n<\omega\}$ is a gamma cover.
Finally, let $\preceq$ denote refinement; then
$$U_{k_n}\in\mathcal U_{k_n}'\subseteq\mathcal U_{k_n}\preceq \mathcal I_{k_n}'\preceq \mathcal I_n'\preceq\mathcal I_n$$
So $U_{k_n}$
is contained in some $V_n\in\mathcal I_k$, so the gamma cover $\{V_n:n<\omega\}$ witnesses property $\gamma'$.
