For the first question, let us consider the following statement:
$x\in\mathbb R$ and $x\ge 0$. It is consistent with this statement that:
- $x=0$,
- $x=1$,
- $x>4301$,
- $x\in (2345235,45237911+\frac{1}{2345235})$
This list can go on indefinitely. Of course if $x=0$ then none of the other options are possible. However if we say that $x>4301$ then the fourth option is still possible.
The same is here. If all sets are measurable then it contradicts the axiom of choice; however the fact that some set is unmeasurable does not imply the axiom of choice since it is possible to contradict the axiom of choice in other ways. It is perfectly possible that the universe of set theory behave "as if it has the axiom of choice" up to some rank which is so much beyond the real numbers that everything you can think of about real numbers is as though the axiom of choice holds; however in the large universe itself there are sets which you cannot well order. Things do not end after the continuum.
That been said, of course the two statements "$\mathbb R$ is countable union of countable sets and "There are non-measurable sets" are incompatible: if $\Bbb R$ is a countable union of countable sets, then there is no meaningful way in which we can have a measure which is both $\sigma$-additive and gives intervals a measure equals to their length; whereas stating that there exists a set which is non-measurable we implicitly state that there is a meaningful way that we can actually measure sets of reals. However this is the meaning of it is consistent relatively to ZF. It means that each of those can exist with the rest of the axioms of ZF without adding contradictions (as we do not know that ZF itself is contradiction-free to begin with.)
As for the second question, of course each set is countable and thus has a bijection with $\mathbb N$. From this the union of finitely many countable sets is also countable.
However in order to say that the union of countably many countable sets is countable one must fix a bijection of each set with $\mathbb N$. This is exactly where the axiom of choice comes into play.
There are models in which a countable union of pairs is not only not countable, but in fact has no countable subset whatsoever!
Assuming the axiom of countable choice we can do the following:
Let $\{A_i\mid i\in\mathbb N\}$ be a countable family of disjoint countable sets. For each $i$ let $F_i$ be the set of injections of $A_i$ into $\mathbb N$. Since we can choose from a countable family, let $f_i\in F_i$.
Now define $f\colon\bigcup A_i\to\mathbb N\times\mathbb N$ defined by: $f(a)= f_i(a)$, this is well defined as there is a unique $i$ such that $a\in A_i$. From Cantor's pairing function we know that $\mathbb N\times\mathbb N$ is countable, and so we are done.
It is a bit hard to answer your first question, since a countable union of countable sets does not even need to be well-orderable. For example, we could have an uncountable set that is the countable union of sets of size $2$. These sets are called Russell cardinals. You may find the following paper interesting:
H. Herrlich and E. Tachtsis, On the number of Russell’s socks or $2 + 2 + 2 + \dots=?$, Comment. Math. Univ. Carolin. 47 (2006), 707-717.
The paper discusses in a fairly self-contained manner the variety of Russell cardinals, and some of their basic properties. As described in the background section of the paper, the name "Russell cardinal" comes from an example due to Russell trying to explain the need for the axiom of choice: If we are given infinitely many pairs of shoes, and asked to pick one of each, we can easily do it since, say, we can pick the left shoes. But if we are given an infinite set of pairs of socks, then it is not clear how to make the choices. Russell's original example, amusingly enough, used boots rather than socks, which kind of makes not much sense, and it is later authors that switched to socks when describing the example.
Anyway, there are many Russell cardinals (at least as many as real numbers!), if there is one at all, and that is just countable unions of pairs.
If we look at countable unions of countable sets, we can have even more possibilities:
- The reals can be a countable union of countable sets. This is a famous example due to Feferman and Levy. On the other hand, the reals cannot be a Russell cardinal, because in any finite set of reals we can always pick one (say, the smallest), so we can easily check that a countable union of finite sets of reals is countable.
- $\omega_1$ can be a countable union of countable sets. In fact, this happens whenever the reals are a countable union of countable sets.
- In a precise sense, there is no bound to the complexity of the sets that can be expressed as a countable union of countable sets. For example, Morris showed that it is consistent to have that for every ordinal $\alpha$ there is an $X$ that is a countable union of countable sets, and there is a surjection from $\mathcal P(X)$ onto $\aleph_\alpha$. (Of course, different $\alpha$ require different $X$.)
This is not to say that there are no limitations. For example, a countable union of countable well-ordered sets has size at most $\omega_1$. It is consistent (by a significant result of Gitik) that any well-ordered cardinal has cofinality $\omega$. This means that $\omega_2$, for example, could be written as a countable union of smaller sets, so as a countable union of countable unions of countable sets. However, it is not itself a countable union of countable sets.
Best Answer
Denumerable means there exists a bijection between the given set and the set $\mathbb {N}$. This indeed, as you point out, creates subtleties for certain proofs. Further to your comment about the countable union of countable sets being countable, it can be shown that without countable choice this result is false. That is, there exists models of ZF where the countable union of countable sets is not countable.