Improvement on the concept of separating families for the union-closed sets conjecture? The union-closed sets conjecture states the following. Let $F$ be a finite family of finite sets that is union-closed and let $\cup (F)$ be the union of all sets in $F$. Then we can find an element in $\cup (F)$ that appears in at least half of the sets in $F$.

One can show that w.l.o.g. we can assume $F$ to be separating, that is for any two elements in $\cup (F)$ we can find a set in $F$ that contains exactly one of the two elements. So in the literature many statements concerning the conjecture are proven for separating families.

However we can do better. Say a family $F$ is given. We now chose an element in $\cup (F)$ and remove it from every set in $F$ if the number of sets doesn't decrease by doing so. We repeat this process as long as we can. After that we obtain a family $\tilde{F}$ with the following property. For every $x \in \cup(\tilde{F})$ we can find $S \in \tilde{F}$ such that $x \in S$ and $S\backslash \{x\} \in \tilde{F}$. We call families with this property minimal. It is easy to see that if $\tilde{F}$ fulfills the conjecture then so does $F$. Thus w.l.o.g. we can assume $F$ to be minimal.

Now we make the observation that being separating is strictly weaker than being minimal. Say $F$ is a minimal family and let $x, y \in \cup (F)$. Then we can find $S \in \cup (F)$ such that $x \in S$ and $S\backslash \{x\} \in F$. If $y \notin S$ then $S$ contains exactly $x$ and if $y \in S$ then $S\backslash \{x\}$ contains exactly $y$. Thus $F$ is separating. Now consider the family $\{\emptyset, \{1,2\},\{2,3\},\{1,2,3\} \}$. One can easily check that it is separating but not minimal. 

This leaves me with the following questions:

Is there any literature about the union-closed sets conjecture 
that introduces a concept like minimal families? 

If not, is an inductive proof for the union-closed sets conjecture plausible
abusing minimal families (reduce the number of sets in a minimal family by removing an arbitrary element from all sets)?

Edit: Here is a follow up.
If I am understanding your idea correctly, I believe you are saying that you can suppose your family doesn't have sets of the form $S$ and $S\cup \{a\}$ for some $a\in U(\mathcal{F})$

You can check Lo Faro's paper "A note on the union-closed sets conjecture" and see that a counterexample must have some sets of that form. Otherwise you wouldn't have a minimal one.
If I am understanding your idea correctly, I believe you are saying that you can suppose your family doesn't have sets of the form $S$ and $S\cup \{a\}$ for some $a\in U(\mathcal{F})$

You can check Lo Faro's paper "A note on the union-closed sets conjecture" and see that a counterexample must have some sets of that form. Otherwise you wouldn't have a minimal one.