Axiom choice, non measurable sets, and countable unions [closed]

I have been looking through several mathoverflow posts, especially these ones http://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice , http://mathoverflow.net/questions/73902/axiom-of-choice-and-non-measurable-set and there still are many questions I would like to ask:

1) According to the first answer of the first post "It is consistent with ZF without choice that the reals are the countable union of countable sets" (and therefore all sets are borel, and hence measurable), however this seems in contrast with the answer to the second post which states that "the existence of a non-Lebesgue measurable set does not imply the axiom of choice" (and therefore it is possible to construct a ZF model without choice where there exists a non Lebesgue measurable set). How can these two statements be both right?

2) I can't understand why the axiom of (countable) choice is necessary to prove that a countable union of countable sets is countable. By saying that the sets are countable, I have already assumed the existence of a bijection from every set to the set of natural numbers, in other words, I have indexed the elements of each set. So what is the problem in chosing elements from each set? This relates to the above topic in that if the AC weren't necessary to prove that countable union of countable sets is countable, then "It is consistent with ZF without choice that the reals are the countable union of countable sets" can no longer be correct, since this would imply that in ZF without choice the reals are countable.

I am only a third year math student with no background in set theory (only naive), so please excuse the ignorance. I hope someone can answer me, thank you!

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This is not really a research-level question; you should try [math.SE](math.stackexchange.com) instead, where I or others will be happy to answer it. – Arturo Magidin Oct 21 2011 at 20:50
I believe that I personally answered the second question a few times on math.stackexchange.com so be sure to read through the relevant tags before posting. It may give you some insight and answers. – Asaf Karagila Oct 21 2011 at 20:55
At least for concern #2, sure, each set is countable, but now we need to know the same for the union of all those sets. You need the power to make some choices. – Christopher A. Wong Oct 21 2011 at 20:58
ok sorry about that...I just posted it on the web site you suggested. Thank you – Emilio Ferrucci Oct 21 2011 at 20:59
1) The existence of a non-measurable set implies "some of the" axiom of choice but not all. 2) enumerable $\neq$ enumerated. – Ramiro de la Vega Oct 21 2011 at 21:10