Consider this with assuming that $\mathcal{F}$ consists of finite subsets of bounded cardinal; let $n$ be the max of these cardinals and let us argue by induction on $n$. The goal is to find $\mathcal{E}\subset\mathcal{F}$ and $U\in\mathcal{U}$ such that $U\subset\bigcup\mathcal{E}$ and and $|E\cap U|\le 1$ for every $E\in\mathcal{E}$.

First, consider a maximal subset $\mathcal{E}$ of the cover such that no element is in the union of others (we call this ``minimal"). So $V:=\bigcup\mathcal{F}=\bigcup\mathcal{E}$. Write $V=V_1\cup V_2$, where $V_2$ is the set of elements of $V$ that belong to at least two elements of $\mathcal{E}$. The minimality of $\mathcal{E}$ implies that no $E\in\mathcal{E}$ is contained in $V_2$ (i.e., has nonempty intersection with $V_1$).

Since the intersection of elements of $\mathcal{E}$ with $V_1$ are pairwise equal or disjoint, we can partition $V_1$ into $n$ subsets each intersecting, each element of $\mathcal{E}$ in at most a singleton. Hence we can conclude if $V_1\in\mathcal{U}$ (with $U$ being one of those $n$ subsets of $V_1$).

Otherwise, $V_2\in\mathcal{U}$. Since $|E\cap V_2|\le n-1$ for all $E\in\mathcal{E}$, we can conclude by induction (and find $U\subset V_2$).

PS Nik Weaver posted an answer while I was writing this one, but I still post, although I guess it's the same idea.