I am reading this article in which two properties of open covers are described:

$\gamma$-property:If $\mathcal U$ is an open $\omega$-cover of $X$, then there sequence $\{ G_n : G_n \in \mathcal U\} \subset \mathcal U$ such that $\underline{Lim} G_n = X$.(page 153)

$\gamma'$-property:If $\mathcal U_n$ is a sequence open $\omega$-covers of $X$, then there sequence $\{ G_n : G_n \in \mathcal U_n \}$ such that $\underline{Lim} G_n = X$.(page 155)$\underline{Lim}A_n=\{ x \in X : \exists n_0 \in \omega \space \forall n \geq n_0 \space x \in A_n \}$

In page 156, it is proved that, $\gamma$-property implies $\gamma'$ property. The general idea of the proof is clear to me except of one remark. It is mentioned at the end of page 155 that, "As we can suppose that $\mathcal U_{n+1}$ is a refinement of $\mathcal U_n$ for every $n \in \omega$, it is enough to prove that there is an infinite subsequence $\langle n_k : k \in \omega \rangle$ and a sequence $G_k \in \mathcal U_{n_k}$ with $\underline{Lim}G_k = X$".

I don't see why we can assume that $\mathcal U_{n+1}$ is a refinement of $\mathcal U_n$ for every $n \in \omega$.

Any help?

Thank you!