Suppose you are given an uncountable, well-ordered set $S$.
Isn't it possible to provide a bijection $f:\mathbb{N} \rightarrow S$
No. $\mathbb N$ is countable. You will end up with a set of elements $\{s_1,s_2,s_3,\dots\}$, but there will exist elements you have not covered. There is nothing in your definition that demands you covered all of the elements.
An example (not a well ordered set, I know, but may still illustrate my point) is if you look at the set $$S=\left\{\frac12, \frac23, \frac34, \dots, \frac{n}{n+1},\dots\right\}\cup[1,2]$$
The procedure you described works on $S$, although $S$ is not well ordered. It takes $\frac12 = s_1$ as it is the least element. Then it takes $s_2=\frac23$ and so on. It produces $s_i = \frac{i}{i+1}$ which is an injection from $\mathbb N$ to $S$, but it does not cover the whole $S$.
Edit: In fact, you can even take the set $$T=\left\{\frac12, \frac23, \frac34, \dots, \frac{n}{n+1},\dots\right\}\cup\{1\},$$
whic is well ordered and is even countable, but your procedure still does not produce a bijection from $\mathbb N$ to $T$.
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
What about $\mathbb{Z}$? That's a nonempty subset of $\mathbb{Z}$; what's its least element?
It is true that every bounded nonempty subset of $\mathbb{Z}$ has a least element (unlike $\mathbb{R}$), but this is a much weaker condition than compactness. (In fact, this is basically trivial in the case of $\mathbb{Z}$ - it's just the statement that bounded sets of integers are finite!)