Baire category theorem (see the section headed "Proof" in that link).
I actually used Baire category theorem. Here goes the solution:
Since , it follows that is a dense subset of F for all lambda. Hence each is a dense subset of F.
Suppose but . That means x is not an adherent point of . Therefore such that . But , so it must be true that .
On the other hand, since each is a dense subset of F, the category theorem guarantees that the countable intersection of these sets is a dense set in F. Thus , a contradiction.