Is the union of finitely many open sets in an omega-cover contained within some member of the cover?

Is the union of finitely many open sets in an omega-cover contained within
some member of the cover?

Let $\mathcal{U}$ be an open cover of $\mathbb{R}$ (Standard Topology)
such that $\mathbb{R} \not \in \mathcal{U}$ and for any finite set $A$
there is a $U \in \mathcal{U}$ such that $A \subseteq U$. We call such an
open cover an $\omega$-cover. Can we show that for any finite set $B
\subset \mathcal{U}$, there is a $V \in \mathcal{U}$ such that $\cup B
\subseteq V$?
Ultimately I'm working on showing the following. Let $\langle
\mathcal{U}_n: n \in \mathbb{N} \rangle$ be a sequence of $\omega$-covers.
Can we find a sequence $\langle F_n: n \in \mathbb{N} \rangle$ with each
$F_n \in \mathcal{U}_n$ such that $\cup F_n$ is an open cover of
$\mathbb{R}$?
My approach here was to use each $\mathcal{U}_n$ to cover $[-n,n]$, thus
eventually covering all of $\mathbb{R}$. Since $[-n,n]$ is compact and
$\mathcal{U}_n$ is a cover, $\mathcal{U}_n$ has a finite subcover. But
that's as far as I can get unless what I conjectured above is true.