Why isn't this a valid argument to the "proof" of the Axiom of Countable Choice? I am having a little trouble identifying the problem with this argument:
Let $\{A_1, A_2, \ldots, A_n, \ldots\}$ be a sequence of sets.
Let $X:= \{n \in \mathbb{N} : $ there is an element of the set $A_n$ associated to $n \}$
(1) $A_1$ is not empty. Therefore, there exists a $x_1 \in A_1$
(2) Given an $A_n$ (and a $x_n \in A_n$), we have that $A_{n+1}$ is non-empty and, therefore, there exists a $x_{n+1} \in A_{n+1}$ 
So, by induction, $X=\mathbb{N}$ and the axiom of countable choice is "proven".
Where is the error?

Imagine, philosophically, mathematical foundation needs to build up a language bridging finiteness at A and countable infinity at B. ACC is such a language leading us jumping from A right to B, while induction is a never ending process leading us to go through each mile in an infinite mileage, but just never get to the end point B.
In a single nonempty set, we can choose an element by the very meaning of the non-emptiness
of the set. Induction then allows us to obtain a choice function for any of the finite collections of sets. Although there are infinitely many of such collections, we come short to obtain a choice function for a countable collection of sets, for that ACC is needed.
In the language of ordinality, induction covers every finite ordinal but comes a little bit short to cover the first infinite ordinal, which is the totality of all finite ordinals, in this sense not accessible by induction.
