Can the axiom of choice be used to define the real numbers? I realize this is possibly a copy of another question (The Axiom of Choice and definability) but I would appreciate an explanation with less set theoretical explanations if that is at all possible.

I recently took the opportunity to look up the axiom of choice (AC). Wikipedia has both a formal and an informal definition.  Based on my shortcomings in set theory I have to rely on the informal version: “given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object from each bin”.

It seems that this can be used to define an endless series of decimals such as an arbitrary number x between 0 and 1, in term corresponding to an arbitrary positive real number through such a function as f(x)=1/x-1, i.e. it can be used to define any real number (assuming a real number is an infinite decimal series).

A beauty spot is the cases of 0.99999… and 1.00000… - discussed in a number of questions on the site. It points to several issues:

I argue, albeit in a roundabout way, that for a choice involving the AC to have any value then the number received through the choice must also allow for existing mathematical operations, i.e. 2*0.11111…=0.22222…  as discussed in another question. (“You're also assuming that you can multiply 0.9999... by 10 and get 9.9999... (or, rather, that arithmetic with infinite decimals works normally), which is not at all unreasonable to assume. – Isaac Jul 20 '10 at 20:59”)
i)  0.99999…must be equal to 1.00000… for the same reason.
ii) The fact that 0.99999…=1.00000…can challenge the one-to-one correlation between the real numbers, as we view as being without these doublets. However it can be disregarded as 

I was assured in a recent question, in that removing a countable set such as 0,99999…, between the endless decimal series and 0.099999…and 0.0099999…  etc. from a uncountable set would not change the one-to-one correlation to numbers without this anomaly. (I was referred to i.a. the Cantor–Bernstein–Schroeder theorem).

Reasoning in this manner I perceive the AC to be identical of postulating - or at least allowing for the definition of – real numbers. My question is whether or not this is the case.

Edit much later: I should add that I primarily am looking at the problems of defing invividual real numbers, so called pointwise definition, rather than definition of the whole set of reals. I didn't realize at the time that these are two different issues.
The axiom of choice has nothing to do with the definition of the real numbers.

We define the real numbers either as Dedekind cuts over the rational numbers, which is simply a pair $(A,B)$ of subsets of the rational numbers which have some properties. Or we can define an equivalence relation on sequences of rational numbers and consider the equivalence classes as the real numbers themselves.

Then we can prove, again without any appeal to the axiom of choice, all sort of things about the real numbers. We can prove basic mathematical facts like convergence or divergence of some geometric series, the definition of $e$ and $\pi$, and $\sqrt2$. We can even prove that none of these are rational. 

Without the axiom of choice we can even prove the Cantor-Bernstein theorem, and we can even prove that every set of real numbers whose complement is countable, has the same cardinality as the real numbers.

No, so far the axiom of choice isn't even playing a role. It does play a role if you want to prove other properties of the real numbers, such as "the real numbers are not the countable union of countable sets" (which is not to say they are countable!). But that doesn't seem to be what you're asking specifically.
It seems to me that you need the Axiom of Choice in order to get most of the Cauchy sequences whose equivalence classes constitute the real numbers.

Consider the set of all rational sequences. There are many which can be defined using a formula, i.e. an algorithm. For example $(1, 1, 1, 1, 1, \dots)$ has the formula $x_n = 1 \ \forall n$. Similarly you can define a rational sequence whose squares converge to $2$ (and thus you get a rational Cauchy sequence whose equivalence class will be named $\sqrt 2$) and thus you have defined that particular real number. It seems to me you can do that to construct all roots of polynomials (i.e. algebraic numbers), but what about all the transcendental numbers? $\Bbb e$ is easy enough. But what about just any old real number which lies somewhere on the number line. How do you define a sequence that converges to it? How can you specify the terms of such a sequence? It seems that you need the Axiom of Choice to do this. Because you're choosing an infinite number of elements from $\Bbb Q$, without any algorithm / formula. 

The point is that to say that infinite sequences (other than ones that can be constructed algorithmically) exist requires the Axiom of Choice. 

It's all very well to say you don't need AoC for constructing the reals because you just need cauchy sequences  - but the Cauchy sequences themselves (most of them - i.e. the ones that can't be specified by using some algorithm) need the Axiom of Choice.