a) Given $x\in\mathbb{R}\setminus\mathbb{Q}$, consider $a_n=\frac{\lfloor nx\rfloor}{n}$. For sure, $a_n\in\mathbb{Q}$.
On the other hand, since $r-\lfloor r\rfloor\in[0,1)$,
$$\left|x-a_n\right|\leq\frac{1}{n}\;\Longrightarrow\,\{a_n\}_{n\geq 1}\to x.$$
b) Given a), any irrational number $x$ can be associated with an increasing sequence of rational numbers converging towards $x$.
c) Since $\mathbb{Q}$ and $\mathbb{N}$ have the same cardinality (by Cantor-Bernstein theorem, for instance), $P(\mathbb{Q})$ and $P(\mathbb{N})=2^{\mathbb{N}}$ have the same cardinality, where $2^{\mathbb{N}}$ is the set of functions from $\mathbb{N}$ to $\{0,1\}$. By considering the binary representation of any number in $[0,1)$, then using the Cantor-Bernstein theorem again, we have that the cardinality of $2^{\mathbb{N}}$
equals the cardinality of $[0,1)$, that equals the cardinality of $(0,1)$ (by Hilbert's paradox, for instance), that equals the cardinality of $\mathbb{R}$ since the map $\varphi:(0,1)\to\mathbb{R}$ given by:
$$ \varphi(x) = -\cot(\pi x) $$
is bijective.