$P_n=\{X\subseteq \mathbb{N}:|X|=n\}$ is countable. Let $n\in \mathbb{N}$ and $P_n=\{X\subseteq \mathbb{N}:|X|=n\}$.
How can I prove that $P_n$ is countable?
Let $(p_i)_{i \ge 0}$ be the sequence of prime numbers with $p_0 = 2$. To each $X = \{x_1, \dots, x_n \} \subseteq \Bbb N$ you may associate a natural number by the formula $X \mapsto p_{x_1} \dots p_{x_n}$. Note that if $p_{x_1} \dots p_{x_n} = p_{y_1} \dots p_{y_s}$ then necessarily $s = n$ and $x_i = y_i$ (modulo a possible reordering). Therefore, the mapping $X \mapsto \Bbb N$ described above is injective, therefore $P_n$ injects into $\Bbb N$, so it is at most countable.

Let us now see that $P_n$ is not finite. Well, this is easy, because it must contain all the subsets of the form $\{k+1, \dots, k+n\}$ for all $k \in \Bbb N$, and these are infinitely many. Therefore, $P_n$ is infinite.

Since $P_n$ is infinite and at most countable, it follows that it is countable.
To show that a set is countable a one to one correspondance of the elements of the set with the elements of N must be  established.
Is countable the union of two countable sets?
Let's see

C ={ c1, c2, c3,...
D ={ d1, d2, d3,...

C U D ={c1, d1, c2, d2, c3, d3,.....cj ,   dj , 
  N   ={ 1,  2,  3,  4,  5,  6,.....2*j-1,  2*j, ...

here we have the one to one rule, between C U D and  N

Now we only have to show that P2 can be expressed as a union of sets, it is isomorphic to N X N = Unions of 1 X N , 2 X N, 3 X N,...
And P3 is built from P2 X N,