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?

The existence of a model of a statement does not mean that statement is "true" (whatever that means; see below). For example, the Poincare disk is a model of Euclid's first four postulates plus the negation of the parallel postulate; this does not mean that the parallel postulate is "false."
What having a model of a set of statements does mean, is: that set of statements is consistent. In particular, if there is a model of $T\cup\{\neg p\}$, then $p$ cannot be a consequence of $T$. Cohen's result, for instance, shows that $AC$ cannot be proved from $ZF$ alone.
As to your comments about its value of an axiom: knowing that there is a model of $ZF$ where $AC$ fails shows that, if we believe that $AC$ is "true" (again, see below), then we have a good reason to adopt it as an axiom. Conversely, if there were no model of $ZF+\neg AC$, that would mean (by Goedel's completeness theorem) that $ZF$ proves $AC$, so there would be no need to add $AC$ as an additional axiom.

OK, so now another question: what is going on when we use the word "true" in these contexts? Basically, we're presupposing the existence of a "correct" model. But the statement "$AC$ is true in the actual universe of set theory" doesn't have anything to do with the statement "there is some model in which $AC$ fails." 
To repeat: all we know from the existence of a model of $ZF+\neg AC$ is that $AC$ is not a consequence of $ZF$.
