I think that I can prove that from one set, you can generate infinitely many sets.
Let A be a closed set of infinite Cantor-Bendixon rank. That is, the successive finite Cantor-Bendixon derivatives A', A'', and so on are all distinct. The rank of an element x in A is the least n such that x becomes isolated in An, where A0 = A and An+1 = A', the set of limit points of An. Thus, the rank 0 elements are the isolated points of A, and the rank 1 elements are the limits of these points which are not limits of limit points of A, and so on.
Now, let B be the subset of A consisting of the elements of A having even rank. This includes all isolated points of A, but not their limits that are not limits of limits, but does include the limits of limits (as long as they are not limits-of-limits-of-limits) and so on. Define the operation B+ = cl(B) - B, which is obtainable from your operations. Note that cl(B) = A, and that B+ consists of all elements of A having odd rank. Similarly, B++ = cl(B+) - B+ consists of all elements having odd rank in B+, which is exactly those elements of A having even rank at least 2. And so on. The set B+n will consist of all elements of A having rank at least n, which have even/odd rank depending on the parity of n. Since A has infinite rank, these sets will all be distinct.