That's a statement given without proof in the following paper (page 12, ch. 3.5) about Frankl's union-closed sets conjecture. It's labeled as easy to prove, but I'm struggling with it. For non-union-closed case the minimum size is at most $(|U(\mathcal{A})|/2) + 1$. Below is a brief overview of the necessary definitions.
Let $\mathcal{A} = \{A_1,\dots,A_n\}$ be a finite familiy of finite sets (in the context, a family of sets just refers to a set of sets). $\mathcal{A}$ is union-closed if $A,B \in \mathcal{A} \Rightarrow A\cup B \in \mathcal{A}$. The universe of $\mathcal{A}$ is $U(\mathcal{A}) = \bigcup\limits_{A \in\mathcal{A}}A$, i.e. the set of all the elements of the sets in $\mathcal{A}$. $\mathcal{A}$ is separating if for any $a,b\in U(\mathcal{A})$ there is $C \in \mathcal{A}$ such that $a \in C, b \notin C$ or $a \notin C, b \in C$.
To prove: the minimum size of a union-closed, separating family $\mathcal{A}$ without an empty set is $|U(\mathcal{A})|$.
The answer to this question can be found in the same paper, as a part of the proof of Theorem 22 (Falgas-Ravry, page 19).
Relabel the elements of $U(\mathcal{A})$ in order of incresing frequency, i.e. so that $|\mathcal{A}_1| \leq |\mathcal{A}_2| \leq \dots \leq |\mathcal{A}_m|$, where $\mathcal{A}_i = \{A \in \mathcal{A} : i \in A\}$. As $\mathcal{A}$ is separating, for each $1\leq i < j \leq m$ there is a set $X_{ij}$ with $i \notin X_{ij}, j \in X_{ij}$. Set $X_i = \bigcup\limits_{j = i+1}^mX_{ij}, 1 \leq i \leq m-1$. Since $i<i' \Rightarrow i' \in X_i, i \notin X_i$, it follows that all $X_i$ are distinct, and also $X_i \in \mathcal{A}$ since $\mathcal{A}$ is union-closed. Together with $U(\mathcal{A})$, they form $m$ distinct sets.