Your claim isn't true.
The positive rationals can be well-ordered
Since $\mathbb{Q}$ bijects with $\mathbb{N}$, the well-ordering on $\mathbb{N}$ will induce a well-ordering on $\mathbb{Q}$ and hence on the positive rationals.
However,
The usual ordering of positive rationals is not a well-ordering
The usual ordering is, of course, $\frac{a}{b}>\frac{c}{d}$ if and only if $ad>bc$ (where $a,b,c,d$ are positive integers).
If it is a well-ordering, then there is a least positive rational $p/q$. But halving it gives a smaller positive rational $p/(2q)$, so $p/q$ can't be the least, contradiction.
Suppose there are $m$ elements in a special set. It seems clear that $m \le \lfloor (N+1)/2 \rfloor$. A well-known result in combinatorics (see below) is that there are $\binom{N-m+1}{m}$ subsets of ${1,2,3,\dots,N}$ of size $m$ with no adjacent elements. Each such set can be ordered in $m!$ ways. So the total number of special sets is
$$\sum_{m=1}^{\lfloor (N+1) / 2 \rfloor} \binom{N-m+1}{m} m!$$
The first few values, for $1 \le N \le 10$, are $1, 2, 5, 10, 23, 50, 121, 290, 755, 1978$. I did a search on OEIS without finding this sequence.
For those not familiar with the formula for the number of ways to select k non-adjacent items from n, here is a derivation. Suppose we want to select $k$ non-adjacent integers $a_1, a_2, a_3, \dots ,a_k$ from the set $\{1, 2, 3, \dots ,n\}$. For the choices to be non-adjacent, we must have $1 \le a_1$, $a_1 < a_2-1$, $a_2 < a_3-1$, $a_3 < a_4-1$, ..., $a_{k-1} < a_k -1$, and $a_k \le n$. An equivalent set of inequalities is
$$1 \le a_1 < a_2-1 < a_3-2 < a_4-3 < \dots < a_k-k+1 \le n-k+1$$
So we may equivalently pick the values $ a_1 , a_2-1 , a_3-2 , a_4-3 , \dots , a_k-k+1$ from $\{1, 2, 3, \dots ,n-k+1 \}$, and this can be done in $\binom{n-k+1}{k}$ ways.
Best Answer
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.