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.
You cannot, in general, ensure that $C(n)(i)$ and $C(n+1)(i)$ are identical for $i \leq n$. If you could do that, you could indeed take $\cup_n C(n)$ to obtain a choice function. But the problem is not with the axiom of union, the problem is that you have overstated what induction actually gives you.
It is true that, when you prove the inductive step, you temporarily obtain $C(n)(i)= C(n+1)(i)$, because you extend the previous finite choice function. But when you apply the principle of induction, all that it tells you is that
$$
(\forall n)(\exists C)[C \text{ is a choice function on } \{A_1, \ldots, A_n\}]
$$
You do not get that there is an infinite sequence $C(n)$ of compatible finite choice functions.
This is one place where the difference between "we construct an infinite sequence inductively" and "we construct a sequence by induction" really matters. To construct an infinite sequence inductively means to give a construction of an infinite sequence $\langle \alpha_0, \alpha_1, \ldots\rangle$ that tells how to construct $\alpha_0$ and then tells how to construct $\alpha_{i+1}$ from $\alpha_i$.
The principle of mathematical induction, on the other hand, has the general form
$$
[P(0) \land (\forall n)[P(n)\to P(n+1)]] \to (\forall n)P(n).
$$
It does not, on its own, give you a way to take a collection of finite sequences and combine them into an infinite sequence. An inductive construction, written in full generality, may require some side proofs by induction to verify that the conditions needed in the construction are maintained at each step. But the power of inductive constructions truly goes beyond the power of mathematical induction.
There are two general situations we can find ourselves in with the inductive construction of a sequence $\langle \alpha_0, \alpha_1, \ldots\rangle$.
Situation 1: We can select $\alpha_{i+1}$ uniquely. For example, there may be a function $f$ so that $\alpha_{i+1}$ can simply be taken to equal $f(\langle \alpha_0, \ldots, \alpha_i\rangle)$. In this case, the axiom of choice is not required for the inductive construction.
One example is the proof of weak Konig's lemma: any infinite $\{0,1\}$ tree has path. We construct the path inductively; given a finite path $\langle \alpha_0, \ldots, \alpha_i\rangle$, if there are infinitely many nodes in the tree above $\alpha_i \smallfrown 0$ then we make $\alpha_{i+1} = \alpha_i \smallfrown 0$. Otherwise we take $\alpha_{i+1} = \alpha_i \smallfrown 1$. This inductive construction does not require the axiom of choice; it can be seen as simply iterating a certain function to produce the path. The function is defined so that $f(\sigma) = \sigma\smallfrown 0$ if there are infinitely many nodes of the tree above $\sigma\smallfrown 0$, and $f(\sigma) = \sigma\smallfrown 1$ otherwise. The axiom of choice is not at all required to construct this $f$.
Situation 2: We can prove the existence of at least one possible value for each $\alpha_{i+1}$, as the construction progresses, but we do not have a function that can be used to guide the construction. This is precisely the situation that the axiom of dependent choice is intended to handle. Dependent choice is precisely the principle you need to prove the axiom of countable choice by inductively constructing the choice function.
Best Answer
The difference is that in the second proof we need to make only two choices: for each of $S$ and $T$ we must choose an injection into $\Bbb N$. Making a finite number of choices does not require any axiom of choice. In the first proof, however, we must make infinitely many choices, since we must choose an injection of each of the sets $S_n$ into $\Bbb N$, and that does require some choice principle.