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?

To avoid Choice, you need to have a definitive way of choosing an element from each set.
For example, if each $A_n$ is a pair of shoes, you may always choose the left. 
A typical useful case is when each $A_n$ has a distinguished member such as a unique minimum that you can choose. For example, Baire Category Theorem for a complete SEPARABLE metric space can be proved by a trick like this.
