How can the axiom of choice be called "axiom" if it is false in Cohen's model? From what I know, Cohen constructed a model that satisfies $ ZF\neg C $.
But if such a model exists, how can AC be an axiom? Wouldn't it be a contradiction to the existence of the model?
Only explanation I can think of, is that this model requires other axioms in addition to ZF, and these axioms contradict AC. Is it true?

You're treating the word "axiom" as you were probably taught in high school. That an axiom is something which is "simply true as an assumption".
Modern mathematics has changed that definition to "an assumption made in a certain context". Not every axiom is called an axiom, some axioms are proved as theorems, and sometimes lemmas are used for axioms. And not to mention various hypotheses used as axioms sometimes.
But in the modern context, an axiom is really just an assumption that you begin with. And different contexts begin with different axioms. From the axioms of $\sf ZF$ we can prove that the axioms of Peano hold in the natural numbers; we can prove that the completeness axiom holds for the real numbers. So these are all theorems of $\sf ZF$.
From the axioms of $\sf ZF$ and Zorn's Lemma we can prove the axiom of choice, so the axiom of choice is a theorem of Zorn's Lemma when working in $\sf ZF$.
When Cohen showed there is a model of $\sf ZF+\lnot AC$, he effectively completed the proof that $\sf ZF$ neither proves nor disproves the axiom of choice. Which is quite fitting as far as "axiom" goes. So you shouldn't worry about it, this is more of a linguistic issue.
