Picking from an Uncountable Set: Axiom of Choice? Question:  Given the real numbers as a set, does it require the (non-finite) Axiom of Choice to pick out an arbitrary single element?  What about if we wanted to pick out an integer?  What about if we wanted to pick out "0"?
Motivation:  Those who have seen some of my previous questions here know that I am terrible at picking out when and where to apply the axiom of choice, and when it is being used in a proof.  Perhaps slightly motivated by the Bertrand Russell quote

"The Axiom of Choice is necessary to select a set from an infinite number of socks, but not an infinite number of shoes."

and perhaps slightly motivated by an argument I had over the AoC, I wanted to find out how much it is actually being used.  The above question is as basic a question as I can think of to test a "lower bound" on the axiom of choice.
 A: Given a finite number of nonempty sets $A_1,\ldots,A_k$, the Axiom of Choice is not needed to show that there are functions $f\colon\{1,\ldots,k\}\to\cup A_i$ such that $f(i)\in A_i$ for each $i$. That is, there is no need for the Axiom of Choice in order to select an element form finitely many nonempty sets.
In particular, you do not need the Axiom of Choice to show that you can choose a real number (a single set).
Simply: since $A_1$ is nonempty, there exists $a_1\in A_1$. Since $A_2$ is nonempty, there exists $a_2\in A_2$. Likewise, we have $a_i\in A_i$, $i=3,\ldots, k$. And we can let $f=\{(i,a_i)\mid i=1,\ldots,k\}$ be the Choice function. We can write all of this down because there are finitely many $A_i$. Apparently there are subtle non-standard-set-theory issues here (thanks to Carl Mummert for the pointer); so instead, let's say that for a family of nonempty sets indexed by a natural number you do not need the Axiom of Choice to get a choice function, and this can be shown by induction on the index set.
The specter of the Axiom of Choice (so to speak) does not even begin to suggest itself until you have to make infinitely many choices. Even then, you may not need it.
Note, however, that using phrases like "arbitrary real number" may make it seem like you are talking about some kind of uniform probability distribution over all the real numbers that makes all "selections" equally likely. This is a completely different thing altogether, but not what you are talking about here. This is likewise the problem that arises when, in the context of probability, we talk about "selecting a random integer." The problem with "selecting a random integer" is that you cannot have a uniform probability distribution on the integers: this would amount to a measure on the power set of the integers that is $\sigma$-additive, for which each integer has equal probability, and for which $\mu(\mathbb{Z})=1$; no such thing exists, because you would need to have $\mu(n)=0$ for each integer $n$, and $\sigma$-additivity would imply $\mu(\mathbb{Z}) = 0$ as well. This is what is behind the statement "you cannot 'choose a random integer'" true in that context; but this is an obstacle in the notion of "randomness", not a set-theoretic obstacle. Likewise, there can be no uniform probability distribution over all the reals, because uniformity would imply that each interval $[n,n+1)$ with $n\in\mathbb{Z}$ would have the same measure, and $\sigma$-additivity together with the fact that the reals have finite total measure implies that each interval must have measure equal to $0$, and so the reals would also have total measure $0$. Again, this is a probability/measure-theoretic obstacle, not an Axiom-of-Choice one.
