(a) Take the lowest element of each subset.

(b) For a subset, take its lowest positive element if there is at least one, otherwise take its largest element (in this latter case the set is bounded from above by $0$ ).

(c) First, for every $q\in\mathbb{Q}$ , we consider its unique representation as a pair of integer numbers $(m_{q},n_{q})$ such that $\frac{m_{q}}{n_{q}}=q$ is irreducible and $n_{q}$ is positive. The fact that it is possible and the pair is uniquely determined should not depend on the axiom of choice. Then, for every nonempty set $A$ of rational numbers, we consider the set of denominators $N=\{n_{q}|q\in A\}\subset\mathbb{Z}_{+}$ , use the choice function defined in (a) to pick one of its elements, i.e. we find the lowest denominator $n=\inf N$ , then, we construct the set $B=\{q\in A|n_{q}=n\}\subset A$ corresponding to the lowest denominator $n$ , consider the set of numerators $M=\{m_{q}|q\in B\}\subset\mathbb{Z}$ , and use the choice function defined in (b) to choose one of its elements, which we call $m$ . Then, $m\in M$ implies $m=m_{q}$ for some $q\in B\subset A$ , for which $n_{q}=n$ , and $\frac{m}{n}=\frac{m_{q}}{n_{q}}=q\in A$ . $q$ is the chosen element.

(d) A nonempty subset of $X^{\omega}$ is a (possibly, uncountable) set of infinite sequences of $0$ and $1$ . We need to provide a general rule that would work for any such subset and pick one of its sequences. Think about it. ;) As a hint, each sequence of $0$ ’s and $1$ ’s corresponds to a real number from $[0,1]$ , and each real number from $[0,1]$ corresponds to one or two (if it is a rational number from $(0,1)$ ) sequences of $0$ ’s and $1$ ’s. If we can explicitly construct a choice function in (d), then, based on it, we can explicitly construct a choice function for the collection of all subsets of $[0,1]$ . But then, one can use this choice function to explicitly well-order $[0,1]$ (see §10), something that is not possible (Feferman?).