Intuition behind the Axiom of Choice Why is it different to make one choice or many choices than to make infinite choices from a theoretical point of view in which indeed you are not going to do any?
How could that be different from making infinite additions for example $\sum_{i=0}^{+\infty}A_n$.
Thanks a lot.
 A: The axiom of choice can be seen as a generalization of the principle of induction. So, since the principle of induction is technically a lot simpler, let's ask why is there a need to accept the principle of induction. 
The principle of induction, for the purposes of this answer, says that for a property $P(n)$ about a natural number $n$, if $P(0)$ holds and if $P(n)$ holds, then $P(n+1)$ holds, then in fact $P(n)$ holds for all natural numbers $n$. 
This principle seems obvious enough (just like the axiom of choice (in at least one of its forms) seems obvious) so why the fuss about calling it a principle? Well, let's first agree that any proof must be a finite list of characters. Now, how does one argue to convince the skeptic about the validity of the principle of induction? One way is to say, well suppose you want to prove that $P(1)$ holds. Then here is a (finite!) proof: $P(0)$ is known. It is also known that $P(0)\implies P(1)$, thus Modus Ponens tells us that $P(1)$ holds. QED.
This of course is far from proving $\forall n\in \mathbb N \quad P(n)$. So we go on. Suppose you want to establish $P(2)$. Well, here is a (finite!) proof: $P(1)$ was already established (i.e., cut and paste prvious (finite!) proof here), and it is given that $P(1)\implies P(2)$. Thus, Modus Ponens again, gives us that $P(2)$ holds. QED. 
Usually one then concludes with the not so convincing argument "and so on" to then argue that we actually established $\forall n\in \mathbb N \quad P(n)$. Well, here is the problem then. We didn't actually prove that! What we did was give a hand-wavy argument that the two assertions 1) $P(0)$ holds and 2) $P(n)\implies P(n+1)$ holds, are sufficient to convince one that one has a recipe for proving $P(n)$ for all $n\in \mathbb N$. In other words, one seems to be convinced that for any given $n$, one can find a (finite!) proof that $P(n)$ holds. But, do we now have a single finite proof that $\forall n\in \mathbb N\quad P(n)$ ? Well, the answer would be yes if you accept the recipe for proofs as an actual proof. In other words, if you accept the principle of induction. 
So, accepting the principle of induction can be said to be the acceptance of a finite recipe of finite proofs for $P(n)$ (where the length of the proof of $P(n)$ depends on $n$ and will typically tend to infinity with $n$) as a single finite proof of all $P(n)$ in one go. It seems very reasonable to accept such a proof recipe as a proof, which is why the principle of induction is doubted by very few. 
Now, the principle of induction is equivalent to the existence of a least element in any non-empty subset of $\mathbb N$, namely to $\mathbb N$ being well-ordered. The axiom of choice, is equivalent to the existence of a well-ordering on any non-finite set. So the axiom of choice allows for more intricate recipes of proofs and is no longer so easily accepted.   
A: Making one choice is simple, if a set $A$ is not empty, then $\exists a(a\in A)$, and therefore we can pick such $a$. This is called existential instantiation. But this choice is completely arbitrary. This is important because in mathematics we are always within the context of writing a proof (even if we only play around, we essentially prepare ourselves for such proof).
However making infinitely many arbitrary choices is something we cannot prove to be possible1. The axiom of choice asserts that we can, in fact, many infinitely many choices at once - as long as we could make each one (i.e. the sets were not empty).
Remember that arbitrary sets have absolutely no structure. We only have $\in$ in our language, and we have sets and their elements. Sometimes we are lucky and the elements of the set are nice enough to allow for a definable way of choosing from them. For example if the empty set is a member or something.
But this need not be the case. The axiom of choice allows us to uniformly endow all the sets with a particular structure from which we can define a selection. 
In comparison, addition of infinitely many real numbers happens within a complete ordered field, where we have some structure, and we use it to establish a criterion when the sum is finite, and if so what is its value.

Footnotes:


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*I am being deliberately imprecise here. The axiom of choice is more than a generalization of existential instantiation for the infinite case. But the intuition which should guide you, in my opinion, is that.
To give a small taste on why things may break down, if we are working within a universe which has non-standard integers then there would be a product which is finite (from the point of view of that model) and therefore is not empty, but since its index set is a non-standard integer we cannot possible write down a formula which instantiate an element from each set.
But all this require first to understand what does internal and external mean in these contexts, and to understand what are non-standard integers and non-standard models better. So it's all pretty far along the road. It is my firm belief that one should start with the idea that the axiom of choice is indeed some sort of a generalization of existential instantiation, and then learn why it is not.
