There is no need to use the axiom of choice here. Suppose $X$ is an infinite well-orderable set. We argue that there is a well-ordered set $(Y,<)$ with $Y$ of strictly larger cardinality than $X$.
For this, consider the set $A$ of all binary relations $R\subseteq X\times X$ such that $R$ is a well-ordering of a subset of $X$. Let's call $X_R$ this unique subset.
We introduce an equivalence relation on $A$ by setting that $R_1\sim R_2$ iff $(X_{R_1},R_1)$ and $(X_{R_2},R_2)$ are order isomorphic. Let $Y$ be the set of equivalence classes.
We can well-order $Y$ by saying that $[R_1]<[R_2]$ iff there is an order isomorphism from $(X_{R_1},R_1)$ to a proper initial segment of $(X_{R_2},R_2)$. One easily verifies that $<$ is well-defined. This means that if $R_1\sim R_3$ and $R_2\sim R_4$, then $(X_{R_1},R_1)$ is isomorphic to a proper initial segment of $(X_{R_2},R_2)$ iff $(X_{R_3},R_3)$ is isomorphic to a proper initial segment of $(X_{R_4},R_4)$. One also verifies easily that $<$ is a well-ordering.
Finally, $Y$ has size strictly larger than $X$. To see this, note first that $X$ injects naturally into $Y$, namely, given a well-ordering $\prec$ of $X$ and any two initial segments $X_1$ and $X_2$ of $X$ under $\prec$, with $X_1$ a proper initial segment of $X_2$, $[\prec\upharpoonright X_1]<[\prec\upharpoonright X_2]$. But each point of $X$ determines an initial segment of $X$.
Now, if there were an injection $f$ of $Y$ into $X$, then the range $Z$ of $f$ would be well-orderable in a way isomorphic to $(Y,<)$ by using $f$ to copy the well-ordering $<$ of $Y$: Simply set $f(a) R f(b)$ iff $a<b$ for any classes $a,b\in Y$. We then have a copy of $(Y,<)$ as a proper initial segment of $(Y,<)$ (again, looking at initial segments of $(Z,R)$), contradicting that $<$ is a well-ordering.
The above may seem complicated but it is simple: All it is saying is that a well-ordering is less than another if it is an initial segment, and this is a "well-ordering of well-orderings". (And we have to use equivalence classes because different well-orderings may actually be isomorphic.)
When $X$ is a countably infinite set, the resulting set $Y$ is well-ordered and uncountable, but any initial segment of $Y$ corresponds to a well-ordering of $X$, so it is countable. If you want a set $B$ as required, simply set $B=Y\cup\{*\}$ where $*$ is some point not in $Y$, ordered by simply making $*$ larger than all the elements of $Y$. Then $\{a\in B\mid a<*\}$ is uncountable, but $\{c\in B\mid c<d\}$ is countable for any $d\in Y$, i.e., for any $d\in B$ with $d\ne *$.
Once you are familiar with the construction of ordinals, the above can be streamlined a bit: Rather than using an equivalence class, we simply use the ordinal isomorphic to any representative of the class. The argument above gives us that the collection of countable ordinals is actually a set, and its union is an (in fact, the first) uncountable ordinal. As a matter of fact, there is not even the need to take a union. The set of countable ordinals is already an uncountable ordinal.
(The argument above shows that given any well-ordered set there is a larger well-ordered set. A similar argument gives that if we have a family of well-orders, we can paste them together to get a well-ordering larger than all the ones in the family.)
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.
Best Answer
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.