How many uncountable subsets of power set of integers are there? The question is to determine how many uncountable subsets of ${P(\mathbb Z)}$ are there. 
I think that the answer is $2^c$.
Let $A=\{B\in P(P(\mathbb Z)):B \text{ is uncountable}\}$
$P(P(\mathbb Z))$ has $2^c$ elements, so cardinality of $A$ is at most $2^c$. 
Of course, I'm having trouble with the lower bound and I'm trying to find an injective function from some set of cardinality $2^c$ into $A$. 
If anybody has any idea, I'd be very grateful!

Start with your definition (after renaming the bound variable) and add a second definition:
$$
A=\{X\in P(P(\mathbb Z)):X \text{ is uncountable}\} \\
B=\{X\in P(P(\mathbb Z)):X \text{ is countable}\}
$$
Consider $f(X) = \overline{X}$ (the complement of $X$ relative to $P(\mathbb Z)$) as a function $f: B \rightarrow A$.  This is a function because every countable subset of $P(\mathbb Z)$ must have an uncountable complement.  Given $X \cup \overline{X} = P(\mathbb Z)$, since the RHS is uncountable, the LHS must also be uncountable, so if $X$ is countable (and in $B$), $\overline{X}$ is uncountable (and in $A$).
The complement is injective by elementary set theory.  Since an injection from $B$ to $A$ exists, $|B| \leq |A|$.  But then we have:
$$
A \cup B = P(P(\mathbb Z)) \\
A \cap B = \emptyset \\
|A| + |B| = |P(P(\mathbb Z))| \\
|A| + |A| \geq |P(P(\mathbb Z))| \\
2|A| \geq |P(P(\mathbb Z))| \\
|A| \geq |P(P(\mathbb Z))|
$$
We already know that:
$$
A \subset P(P(\mathbb Z)) \\
|A| \leq |P(P(\mathbb Z))|
$$
So the equality follows by Schröder–Bernstein.
