I have a fake proof of the axiom of countable choice. Obviously it is not correct, but I cannot see my flaw. Forgive me, I am only learning set theory.
Let $\{A_n : n \in \mathbb{N}\}$ be a countable collection of nonempty sets. Since the axiom of finite choice is a theorem in ZF, there is a choice function $f_n : \{A_n\} \to A_n$ for every $n$. Let $f = \bigcup_{n \in \mathbb{N}} f_n$. Then $f$ is a choice function on $\{A_n : n \in \mathbb{N}\}$.
Unless each $A_n$ is a singleton (or at least all but finitely many are singletons), there are many [read: more than one] choice functions from each $A_n$. This means that you had to choose each $f_n$, which is precisely where you appeal to the axiom of countable choice.
For example, if $A_0=\{x,y\}$ do you take $f_0$ to be the function which chooses $x$ or the one which chooses $y$?