I recognize there are posts already generally on the logical equivalence of induction and the well-ordering principle, however, I'd really appreciate some advice for finding the monster lurking underneath this marsh of poor reasoning. Thanks!
Complete induction $\implies$ well-ordering principle
Consider a statement $S$ where $S(n)$ states that a given subset of $\mathbb{N}$ with cardinality of $n$ has a least element.
Let, for a set $X \subseteq \mathbb{N}$, $|X| = 1$. By the reflexive axiom, $\forall a \in X, a \le a$ so $S(1)$ holds.
Now we assume $S(n)$ is true so $|X_{n}| = n$ and $\exists \ a\ \forall \ b\ : (a,b \in X) \implies a \le b$.
$S(n+1)$ would state that the well-ordering principle holds for a set $X_n \cup {z}$ where $z$ is a new element.
Since the well ordering principle held on $X$, there is a least element of $X$, call it $a$. Now $(z \le a)\vee (z > a) $. If the former is true than $z$ is now the least element. If the latter is true then $a$ is still the least element.
Why is this seemingly much more obvious "solution" wrong? I have a gut feeling this has something to do with considering arbitrary elements of the power-set of $\mathbb{N}$ - which is uncountably infinite - and forming a bijection from $\mathbb{N}$ to these subsets of $P(\mathbb{N})$.