In classical logic every formula is equivalent to a formula in prenex normal form (i.e., a formula starting with a string of quantifiers followed by a string of terms called the matrix of the formula). The Wikipedia link above details the conversions for disjunction, as is the case here. You have (x)(x not in A or (z)(x not in z or x in A)). This is equivalent to (x)(x not in A or (z)(x not in z) or x in A) which is then equivalent to (x)(z)(x not in A or x in A or x not in z), as you stated. However, I fail to see how this represents A is a subset of the power set of A. In fact, it is just not true that A is a subset of the power set of A. Take a simple counter example. Let A = {1, 2}. Let 0 denote the empty set. Then the power set of A is {0, {1}, {2}, A}. Now it is true that A is in the power set of A, but to be a subset of it A must contain something of the power set. In this case, that would be like saying A = {1, 2, {1}} or A = {1, 2, A} or {1, 2, {}} = {1, 2, 0}. Therefore, it is not true. Both the antecedent and the consequent of that implication are false. However, classically this entails that it is a valid material conditional because a false antecedent makes it vacuously true.

I have to question, though, whether or not your proof was to be logical or mathematical. You approached it by claiming some logical statements. The mathematician, however, would follow a typical methodology. To prove the implication you first assume the antecedent. The goal is to then derive the consequent. To derive the consequent you need to show that p(A) is a subset of its power set. To show something is a subset you arbitrarily choose an element, call it x, from p(A) and then demonstrate that it also is in p(p(A)). Since the choice was arbitrary, it could have been any element in p(A). Therefore, we can universally generalize this result to all x (universal quantifier introduction). This establishes the proof (whether or not the antecedent assumed was required in the proof). But as I said, the terms of this implication are false. You will not be able to show for any x in p(A) that it is also in p(p(A)).