Challenge 9: an element of density at least $\tfrac12-\tfrac1{r+2}$ in every union-closed family

For a natural number $r \in \{0, 1, 2, \dots\}$, $\operatorname{statement}_{09}(r)$ is the following proposition (it is additionally parametrised by an arbitrary, fixed universe in which the type $U$ below lives).

For every type $U$ equipped with a decidable equality (a procedure deciding whether two elements of $U$ are equal; $U$ may be finite or infinite, and is otherwise unrestricted), and for every finite family $F$ of finite subsets of $U$ (a finite set whose members are finite subsets of $U$, without repetition), if

1. $F$ is union-closed: for all finite subsets $A, B$ of $U$, $A \in F$ and $B \in F$ imply $A \cup B \in F$ (with $A = B$ allowed; nothing is required about intersections or about $\emptyset$), and
2. $F$ is nondegenerate: $F \neq \emptyset$ and $F \neq \{\emptyset\}$ (literally equivalent to: $F$ has at least one nonempty member; $\emptyset \in F$ is allowed, and $F$ may consist of a single set),

then there exists an element $x \in U$ such that

Here $|\cdot|$ counts members, so $\operatorname{dens}_F(x)$ is the number of members of $F$ containing $x$, divided by $\max(1,|F|)$ (natural numbers viewed as rationals; the maximum is taken among natural numbers). Under hypothesis 2 one has $F \neq \emptyset$, so the denominator is exactly $|F|$ and $\operatorname{dens}_F(x) \in [0,1]$ is the fraction of members of $F$ that contain $x$. On the right-hand side $r$ is viewed as a rational number, $r + 2 \ge 2$, so no division by zero occurs; the bound equals $\frac{r}{2(r+2)}$, which is strictly less than $\frac12$ for every $r$ (it is $0, \frac16, \frac14, \frac{3}{10}, \dots$ for $r = 0, 1, 2, 3, \dots$). The inequality is non-strict ($\ge$).

Quantifier structure and edge cases:

• $r$ is fixed before $U$ and $F$ are chosen; the witness $x$ may depend on $U$ and $F$ (and on $r$). Only existence is asserted, not uniqueness, and $x$ is not required to belong to any member of $F$.
• For $r = 0$ the required bound is $0$; since $\operatorname{dens}_F(x) \ge 0$ for every $x$, the conclusion then only demands that some element $x$ of $U$ exists.
• If $U$ has no elements, its only finite subset is $\emptyset$, so $F$ is $\emptyset$ or $\{\emptyset\}$ and hypothesis 2 fails: the statement holds vacuously for such $U$. More generally hypothesis 2 forces $F$ to contain a nonempty set, hence $U$ to have an element.
• The hypotheses are satisfiable (e.g. $F = \{\{a\}\}$ for any $a \in U$), so the statement is not vacuous overall.