Where does this proof that $P(\mathbb{N})$ is countable fail?

Question

Let $A_k$ be the set of all subsets of $P(\mathbb{N})$ with size $k$.

$A_k\in\mathbb{N}^k$ is countable for $k\in\mathbb{N}$.

Since $$P(\mathbb{N})=\bigcup_{k=1}^\infty A_k$$ which is a countable union of countable sets, $P(\mathbb{N})$ must also be countable.

However, $P(\mathbb{N})$ is known to be uncountable. Which step of the proof is wrong?

Answer 1

This shows that the set of all finite subsets of $\mathbb{N}$ is countable. This does not work for the full powerset. $\mathscr{P}(\mathbb{N})$ contains infinite subsets, for instance the subset of even integers is not contained in your union but is a subset of $\mathbb{N}$.

Answer 2

Not every subset of $\Bbb{N}$ is finite, so the wrong step is writing the powerset as the union of the collection of all finite sets of size $k ≥ 0$.

Answer 3

$$P(\mathbb{N})=\bigcup_{k=1}^\infty A_k$$

Is not true, as there are infinite subsets of $\mathbb{N}$ which are not included in the union above. Alternatively, if you interpret that union to include $A_{\infty}=$'the infinite subsets of $\mathbb{N}$', then that collection is not countable, hence you cannot use "countable union of countable sets is countable" to complete the proof.