Let $A$ be a set and $F\subset \mathcal P(A)$, and for any $a\in A$, let's define $F_a:= \left\{X\in F, a\in X\right\}$
We suppose that
1)There is no chain $C\in F$ such that $\bigcup (F-C)\subset \bigcap C$
2)For any $E\subset F$, $\bigcap E\in F$
If there exists $a\in A$ such that $|F_a|\leq |F-F_a|$ then let's say $F$ has the Frankl property, or that $a$ is rare (in $F$).(*)
The Frankl conjecture is equivalent to the statement that any $F$ that has finite cardinality has the Frankl property. (well it says that $|F_a|<|F-F_a|$ because condition 1) implies $A\notin F$ but we can show that Frankl conjecture is true iff (*) for any finite $F$ that satisfies 1) and 2), but this has no importance)
The reason why I keep (*) this form is the question :
Is there an infinite $F\in \mathcal P(A)$, satisfying 1) and 2) but such that there is no rare elements ?
First observe that the trivial counterexample to most of the attempt of generalisation, that is an infinite chain, does not satisfy 1)
I would have, if there is no counterexample were found, another question : could it be possible to deduce the truth (resp. the falsegness) of this generalisation for infinite, from the truth (resp. falseness) of Frankl Conjecture? (despite it is not a first order problem)
Let $A=\mathbb{N}$ and let $F$ be the set of subsets of $\mathbb{N}$ that have either the form $\{0,\dots,n-1\}$ or the form $\{0,\dots,n-1,n+1\}$ for some $n\in\mathbb{N}$. This satisfies your conditions (1) and (2) (for (1), note that any chain $C$ can only contain one of $\{0,\dots,n\}$ and $\{0,\dots,n-1,n+1\}$ for each $n$ and so $\bigcup (F-C)$ must be all of $\mathbb{N}$). However, every element of $\mathbb{N}$ is contained in all but finitely many elements of $F$, so no element is rare.