You don't need the Axiom of Choice here to pick a rational in each $(a-\frac{\alpha}{2}, a+\frac{\alpha}{2})$. You can fix an enumeration $(q_r)$ of the rationals and just pick the rational $q_i \in (a-\frac{\alpha}{2}, a+\frac{\alpha}{2})$ with the smallest index $i$.
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.
Best Answer
No, there is no actual use of choice. We can always assume the distance is rational, since once $B_r(x)=\{x\}$, then for every $r'<r$, it is also true that $B_{r'}(x)=\{x\}$, so in particular we can assume that $r$ was a rational number. Since the rationals are countable, we can enumerate them and choose the least one (in the enumeration!) that works.
And since $A$, the dense set, is countable, we can again enumerate it and choose the least one.
So, while the phrasing is one that relies on choice, in actuality, we can easily avoid it.