If I'm not wrong, it is easy to prove the following statement:

If $n \leq 4$ is a natural number, if $\mathcal{F}$ is a union-closed family of non-empty sets, if the universe of $\mathcal{F}$ (i.e. the union of all members of $\mathcal{F}$) has exactly $n$ elements, if $\mathcal{F}$ is separating (i.e. for any two distinct elements $x, y$ in the universe of $\mathcal{F}$, there is a member of $\mathcal{F}$ that contains exactly one of $x, y$), then $\mathcal{F}$ has at least $n$ members and if it has exactly $n$ members, these members have a common element.

Is there a more general statement (for $n \geq 5$) in the literature ? Thanks in advance for the answers.