Let
be the power set of
.
We show that given
, either
or
. Indeed, both
and
cannot be in
because their intersection is empty. Thus the map
which maps
is well-defined, and a bijection because
contains
elements. Since the domain and the image of this map partition
, one of
or
must lie in
.
Now suppose
is not closed under intersection. Then there exists
such that
. By the above, we must have that
is in
. But
, which contradicts the hypothesis that any three elements of
have a nonempty intersection. Therefore
is closed under intersection, and
; since
does not contain
we are done.