Understanding axiom of choice from examples

Solution 1:

In fact $\mathsf{AC}$ is unnecessary in each of the examples in your post (before the edit that added example 7). I'll explain $5B$; in fact, the same choice function works for the other examples you ask about.

(I'm assuming that the "$\le$"s should be "$<$"s when open intervals are concerned, since $(a,a)=\emptyset$ and that has nothing to do with choice.)

The key point isn't that $\mathbb{Q}$ is countable per se, but rather that it is well-orderable (which is weaker). Specifically, fix at the outset a bijection $f:\mathbb{N}\rightarrow\mathbb{Q}$. Now given $a<b$ let $$c_{a,b}=f(\min\{n: f(n)\in (a,b)\}).$$

It's easy to see that $c_{a,b}\in(a,b)\cap\mathbb{Q}$ whenever $a<b$. Consequently the map sending the interval $(a,b)$ to the number $c_{a,b}$ is - modulo appropriate choice of $f$ - a "simply-definable" choice function for the family of intervals in $5B$, and so choice plays no role.