If there are only two ideals, they don't even have to be prime. If H contains x from P1-P2 and y from P2-P1, H must contain x+y, which is not contained in either ideal. Yet all the members of H come from some ideal, hence H belongs to one of the two ideals.
For more than two ideals we need primality. Proceed by induction on the number of prime ideals.
Suppose we have a minimal counter example; the smallest number of prime ideals that refutes this theorem. If H is contained in some of the prime ideals, but not all, it is contained in one of them by induction. Thus a piece of H is in every prime ideal, and outside of the others.
Let xi be an element in H that comes from Pi, and is not present in the other prime ideals. For each i, let yi be the product of xj, where j is not equal to i. In other words, yi is the product of the other x values. Let z be the sum of yi. For every prime ideal Pi, all the terms of z are contained in Pi, except for yi. Thus z is in Pi iff yi is in Pi. Now z is in H by closure, so let it lie in the ith prime ideal. Thus yi is in Pi, and yi is a product, and represents the product of principal ideals in our commutative ring. Thus some xj from that product is in Pi, and that is a contradiction. Therefore the set H is always contained in one of the prime ideals.