Yes!
First and foremost, although the power set operation gives a larger cardinality, it need not give the next larger cardinality (I'm assuming that the cardinals are well-ordered here, which follows from ZFC). This is the heart of the Continuum Hypothesis: the reals are essentially the power set of the naturals, are there sets of reals of intermediate cardinality?
Second of all, we have the notion of limit cardinals. The following is a chain of sets of increasing cardinality:
$$\aleph_0, \mathcal{P}(\aleph_0), \mathcal{P}(\mathcal{P}(\aleph_0)), \dots$$
But what's the cardinality of their union? It's bigger than all of them, but it's not equinumerous with the power set of anything smaller than it. (Can you see why?)
The remaining question is, can we climb up past all the cardinals via repeated applications of power set and union? This leads to the notion of inaccessible cardinals. These are cardinals $\kappa$ which are larger than the power set of anything smaller cardinal $\lambda < \kappa$, and larger than any union of less than $\kappa$ many sets each of which has cardinality less than $\kappa$. That is:
- If $\lambda < \kappa$ and $\mu_i < \kappa, \forall i < \lambda$, then $\left|\bigcup_{i<\lambda}\mu_i\right| < \kappa$
The existence of such inaccessible cardinals is not provable from ZFC (unless ZFC is inconsistent). In fact, the hypothesis "there exists inaccessible cardinals" is so strong that ZFC + "there exist inaccessibles" implies the consistency of ZFC! This is of course strictly stronger, because ZFC alone cannot prove its own consistency, a la Godel. Inaccessible cardinals are one type of the famed large cardinals.
The set you are talking about is sometimes written as $\def\N{\Bbb N}\binom\N2$ in combinatorics: the set of all $2$-element subsets of$~\Bbb N$. It has the same cardinality as$~\N$. An explicit bijection is given by the combinatorial number system system for $k=2$: map the subset $\{a,b\}$ with $a<b$ to $\binom a1+\binom b2=a+\frac{b(b-1)}2$.
A similar bijection holds for $k$-element subsets for any fixed finite$~k$ (see the linked page). Even the union of all those sets is in bijection with$~\N$; in this case a slightly different bijection works: interpret the subset as a binary number, with the bit in position$~i$ indicating whether $i$ is present in the subset (value $1$) or not (value $0$).
In order to get an uncountable collection of subsets of$~\N$, your collection must contain some subsets that are infinite, and moreover need infinite information to describe (a necessary but not sufficient condition). As long as each element can be specified, using some fixed and definite encoding, by a finite amount of data, one can order the elements by size of those data first, and then for a fixed size by some total ordering of the encoding data (lexicographically for instance); this produces a well-ordering in which each element occurs in a position corresponding to a natural number. The above bijections are in fact inspired by fairly simple such ordering schemes, but using "largest element" in the place of "data size" (but that would no longer work for collections containing infinite subsets with finite descriptions).
Best Answer
The cardinality of the union is exactly the cardinality of $\mathbb R$. To see that, take the extreme case that $A$ and $B$ are disjoint and use the bijections $f:A\to R$ and $g:B\to \{0,\dots,k-1\}$ where $k=\lvert B\rvert \in \mathbb{N}$ to construct a bijection $\phi: A\cup B\to\mathbb{R}$ as follows: $$\phi(x) = \cases{ g(x) & for $x\in B$\\ f(x)+k & for $x\in A$ and $f(x)\in\mathbb{N}$\\ f(x) & for $x\in A$ and $f(x)\notin\mathbb{N}$ }$$ If $A\cap B\neq\emptyset$ then use the fact that $A\cup B=A\cup(B\setminus A)$ and do the same construction with $B\setminus A$ instead of $B$ (of course, then $k=\lvert B\setminus A\rvert$).
Note that a similar construction works also if $B$ is countable infinite. In that case, map $B$ to the odd numbers and replace $f(x)+k$ with $2f(x)$ for $f(x)\in \mathbb N$.