A. is correct; it's one of the paradigm wellordered sets.
C. is not correct. This has no least member, since for any $x>0$ one can take $x/2$ and get a smaller number greater than $0$, and hence a positive rational.
I'm not sure what you mean by "inversing", but the inverse order has no direct bearing on whether something is wellordered. The only thing that matters is if there are any subsets without a least member. For instance, the integers are not wellordered because $\mathbb{Z}$ itself has no least member; the negative numbers keep going down forever. Likewise the positive rationals keep getting smaller and smaller forever.
To show that a set is not wellordered, you just need to show a non-empty subset such that for every $x$, there's a $y$ with $y<x$. To show that it is wellordered, you need to show that there are no such sets.
You have to re-read all the passage from page 67-on :
We shall say that a well ordered set $A$ is a continuation of a well ordered set $B$, if, in the first place, $B$ is a subset of $A$, if, in fact, $B$ is an initial segment of $A$, and if, finally, the ordering of the elements in $B$ is the same as their ordering in $A$.
Thus if $X$ is a well ordered set and if $a$ and $b$ are elements of $X$ with $b < a$, then $s(a)$ is a continuation of $s(b)$, and, of course, $X$ is a continuation of both $s(a)$ and $s(b)$.
If $\mathcal C$ is an arbitrary collection of initial segments of a well ordered set,
then $\mathcal C$ is a chain [see page 54 : "a totally ordered set"] with respect to continuation; this means that $\mathcal C$ is a collection of well ordered sets with the property that of any two distinct members of the collection one is a continuation of the other.
A sort of converse of this comment is also true and is frequently useful. If a collection $\mathcal C$ of well ordered sets is a chain with respect to continuation, and if $U$ is the union of the sets of $\mathcal C$, then there is a unique well ordering of $U$ such that $U$ is a continuation of each set (distinct from $U$ itself) in the collection $\mathcal C$.
Roughly speaking, the union of a chain of well ordered sets is well ordered. This abbreviated formulation is dangerous because it does not explain that "chain" is meant with respect to continuation. If the ordering implied by the word "chain" is taken to be simply order-preserving inclusion, then the conclusion is not valid.
The relevant fact is : the collection $\mathcal C$ of well-ordered set is a chain w.r.t continuation.
A collection $\mathcal C$ is a chain when, for all $A,B \in \mathcal C$ : $A \subseteq B$ or $B \subseteq A$.
If a collection $\mathcal C$ is a chain w.r.t continuation, it has "something more" : in addition to the property (common to all chains) that for all $A,B \in \mathcal C$ : $A \subseteq B$ or $B \subseteq A$, we have also that (supposing : $B \subseteq A$) $B$ is an initial segment of $A$, and the ordering of the elements in $B$ is the same as their ordering in $A$.
Thus, when we "merge" all the members of the collection $\mathcal C$ into the "mega-set" $U$ every subset "preserve" its "original" minimal-element.
I hope it may help ...
I'm not able to "manufacture" an example different from the "trivial" one built from $\mathbb N$.
If you consider a collection $\mathcal C = \{ X_n \}$ where all $X_i$ form a chain w.r.t continuation, we have that $X_n = \{ 0,1,2, \ldots n \} = n+1$.
Thus the union $U$ of the collection is $\omega$ itself.
Best Answer
The following proof assumes the following is known:
We have to show that every nonempty subset $A$ of $S$ has a minimal element. As $A\ne \emptyset$, let $a\in A$ be some element of $A$. Then $B:=\{x\in A\mid x\le a\}$ is nonempty and finite because it is in fact $\mathbb Z\cap [r,a]$. Hence $B$ is well-ordered. Let $b=\min B$. Then $b=\min A$ because for $x\in A$ we have either $x\le a$ and then $b\le x$ because $x\in B$; or we have $x>a$ and then $x\le a<x$ directly.