I am trying to prove the following proposition:
Proposition: Let $S$ be a countable set. Then the set of all finite subsets of $S$ is also countable.
My approach:
If $S$ is countable that means that $S$ is finite or that $S \sim\mathbb{N}$.
If $S$ is finite then the number of finite subsets of $S$ is also finite, so in this case it's easy to show that this is countable as well.
In the case that $S$ is countably infinite:
Let $A_k=\{A \subset S:\text{card} (A)=k\}$. $A_k$ is countable because it is the union of various countable sets $A \subset S$
The set of all finite subsets of $S$ will be equal to $\bigcup \limits_{i=1}^\infty A_i$. Because every $A_i$ is countable, then this union is also countable.
My question is: Can I do this union? Because $A_\infty$ means the set is the set of all infinite subsets of $S$, and I'm making an union from $i=1$ to $\infty$. So is this proof right?
There is no problem with that union. However, it is wrong to assert that $A_k$ is countable “because it is the union of various countable sets”. An arbitray union of countable sets doesn't have to be countable. You have to justify that you have a countable union of countable sets here.