In order to prove that $M=\{\text{finite subsets of}\; \mathbb{N}\}$ is countable (I know, this question appeared many many times on mse), I want to define a surjective map $f:\mathbb{N}\to M$. I am aware of that there are many other proofs already available on mse!
My guess for such a map f is $f(n)=A_n$, where $A_n$ is a subset of $\mathbb{N}$ with $n$ elements, but I'm not sure if this is correct (since I don't know what happens for n very large, if this is well-defined). Can I choose $f$ like this, or how to define a surjective map $f:\mathbb{N}\to M$?
Correction: What about $f(n)=\{1, 2,\cdots, n\}$, does it work?
No. For example $f(2)=\{1,2\}=\{3,4\}=\dots=\{a,b\}$ which means that $f$ is not a function at all.
If you know that a countable union of countable sets is again countable, then one can proceed in the following way:
Let $A_k$ be the collection of sets with cardinality $k$ (as you do.) Instead of setting $f(k)=A_k$, instead consider $f_k:A_k \to \prod_{i=1}^{k}\mathbb N$ which maps each element of $A$ to a a tuple, arranged by the $<$ relation. Note that $|A_k| \leq |\mathbb N|$.