One of the equivalent statements of the Axiom of Choice is that
$\prod X_i\neq \emptyset\iff X_i\neq \emptyset\text{ for all }i.$
One can prove that this is equivalent to the statement that
Every chain in every partial order is contained in a maximal chain.
But why you have to assume AC for that? If you have a chain you can just add every element which is compareable to every element of the chain, to get a new chain, right?...
I guess this intuitive argument works only for countable sets, where you can really take new elements "step by step" and add them to your chain. But with AC we can make sure that its also working for arbitrary amount of sets i.e. with AC we can guarantee that the "thing" we will get after adding all the elements is indeed a chain. Am I right? It is still a bit confusing for me. I hope that someone can clarify a bit...
The problem with your suggestion is that it only works finitely many steps. There are two problems here:
Usually there is no canonical choice as to what element you're adding to your chain, and this may affect your future choices as well (if you add $x$, then you cannot add any element incomparable with $x$ later on).
The reasoning "Well, we can just do one more step" is useful for induction on the natural numbers, which tells you---essentially---that you can do something for every finite number of iteration. But an induction on the natural numbers does not even let you do something infinitely many times. For this you need the axiom of choice.
So the axiom of choice is used to ensure that we can make coherent choices, and that we can iterate them infinitely many times if need be.