One more strengthening of Frankl's conjecture If I'm not wrong, it is easy to prove the following statements :
1° For every natural number $k$, every union-closed family that has at least $2$ members with cardinality $\geq k$ has at least $2$ members whose intersection has cardinality $\geq k$. This is trivial : if $X$and $Y$ are two distinct members of the family  with cardinality $\geq k$, we can assume that $Y$ is not contained in $X$. Then $X$ and $X \cup Y$ are two distinct members of the family whose intersection is $X$ and has thus cardinality $\geq k$.
2° For every natural number $k$, every union-closed family that has at least $k+3$ members with cardinality $\geq k$ has at least $3$ members whose intersection has cardinality $\geq k$.
3° For every natural number $k$, every union-closed family that has at least $k+4$ members with cardinality $\geq k$ has at least $4$ members whose intersection has cardinality $\geq k$.
I think I have also a proof of the following statement :
4° For every natural number $k$, every union-closed family that has at least $\frac{k^{2}+5k}{2} + 5$ members with cardinality $\geq k$ has at least $5$ members whose intersection has cardinality $\geq k$.
My question is : are there results such as : "If a union-closed family of sets has at least this number of members with cardinalty $\geq k$, then the family has at least $c$ members whose intersection has cardinality $\geq k$".
(Edit :) For every natural number $c \geq 1$ and for every natural number $k \geq 0$, let $s(c,k)$ denote the least natural number $s$ with the following property : every union-closed family that has at least $s$ members with cardinality $\geq k$ has at least $c$ members whose intersection has cardinality $\geq k$. (I presume that such an $s$ always exists, so that $s(c,k)$ always exists.) Clearly, $s(1,k) = 1$ and the above statements amount to say that $s(2,k) \leq 2$, $s(3,k) \leq k + 3$,  $s(4,k) \leq k + 4$, $s(5,k) \leq \frac{k^{2}+5k}{2} + 5$. In fact, it is easy to find examples proving that $s(2,k) = 2$, $s(3,k) = k + 3$,  $s(4,k) = k + 4$, $s(5,k) = \frac{k^{2}+5k}{2} + 5$.
(In order to prove that $s(5,k) \geq \frac{k^{2}+5k}{2} + 5$, look at the family of all subsets of a set with cardinality $k+2$.)
Thus, it seems reasonable to conjecture that for every natural number $c \geq 1$, there is a polynomial $f_{c}(X)$ such that for every natural number $k$, $s(c,k) = f_{c}(k)$.
The first poynomials of the sequence are  $f_{1}(X) = 1$, $f_{2}(X) = 2$, $f_{3}(X) = X+3$, $f_{4}(X) = X+4$, $f_{5}(X) = \frac{X^{2}+5X}{2} + 5$.
Is there a known sequence of polynomials beginning in this manner ?
Note : for every natural number $r \geq 0$, we must have
$s(1 + 2^{r}, k) \geq {k+r \choose k} + {k+r \choose k+1} + \ldots + {k+r \choose k+r} +1$.
(Look at the family of all subsets of a set with $k+r$ elements.)
In the limits of the above results, $s(1 + 2^{r}, k)$ is equal to the found minimal value, $s(c, k)$ grows unit by unit when $c$ runs from $c = 1 + 2^{r}$ to $c=2^{r+1}$ (included) and, for $c=1+2^{r+1}$, takes again the found minimal value (with $r+1$ instead of $r$).
 A: I do not know of a reference for your version.  I provide some motivation for recasting your question into lattice terms.
Along similar lines is the following lower bound argument:
Assume there is a counterexample to Frankl's conjecture.  Thus there is a collection F which contains its common nonempty union X.  Since X is nonempty, there is an element z which occurs in X meaning it occurs in at least one member of F. Thus F (being a counterexample) has at least 2*1+1=3 members.
Since F has at least 3 members, two of the members are nonempty, and either one of them contains the other, or their union contains both.  Either way, we find an element z in at least two members of F.  So F contains at least 5 members.
Now we have at least 4 nonempty members of F to use. We assume without loss a nonempty member A of F contained in a nonempty member B of F. If third member C intersects A or A union C is distinct from B, we have z in A also in two other members of F.  Otherwise B is the disjoint union of A and C and now we use D distinct from C to get D or D union A a third set containing either z or else B, C and D having a nonempty intersection. This means F has at least 7 members.
One can pursue this line with increasing complexity up to F having at least 13 members (I might remember/resurrect the argument from old notes); I am essentially looking at (semi-)lattices and trying to prove an equivalent inequality regarding the number of join-irreducible elements in the lattice.  You might find some benefit in recasting your problem in lattice-theoretic terms.  Poonen has shown such an equivalence in his 1990 REU paper, and by now the surveys on Frankl's conjecture should include this version and some progress on it.
I will look for my notes on derivations and post some related ideas, e.g. if union-closed finite F has k-many 2 sets then F has at least as many (-2)sets (shorthand for $|X|-2$), or F is one of finite and small number of exceptions.
Gerhard "Is Hobbling Down Memory Lane" Paseman, 2016.03.31.
