Well ordering on the quotient of well ordered sets Let $X\neq\emptyset$ be some set. Consider $\mathcal A$ to be the set containing all pairs $(Y,\leq)$, where $Y\subseteq X$ and $\leq$ is a well-ordering. Define
$(A,\leq)\equiv (B,\leq')$ iff $(A,\leq)$ is order isomorphic to $(B,\leq')$, which is clearly an equivalence relation. Further, define, in $\mathcal A/\equiv\,= :\hat{\mathcal A}$ an ordering 
$[(A,\leq)]\preceq [(B,\leq')]$ if there exists an initial segment of $B'\subseteq B$ such that $(A,\leq)\equiv (B',\leq')$. The ordering is strict if all $B'\subset B$ are proper initial segments.
I have verified this ordering is well-defined and is a total ordering since for any two well ordered sets there is an order isomorphism from one to an initial segment of the other.
Now, fix $(A,\leq)\in\mathcal A$ and consider the set 
$$\hat{\mathcal S}_A := \{ [(B,\leq')] : [(B,\leq')]\prec [(A,\leq)]\} $$
The mapping $f: (A,\leq)\to \hat{S}_A, x\mapsto [(A_x,\leq)]$, where $A_x := \{z\in A: z<x\}\subset A$ is a proper initial segment, is an order isomorphism.
From the above, we are to deduce the set $(\hat{A},\preceq)$ is actually well-ordered.
To finish A. Karagila's post. If $B$ is not least in $\mathcal B$, then $B'\in\mathcal B$ and $B'\prec B$, hence $B'$ embeds into an initial segment $A_y\subset A$, but then $y\in A$ to which corresponds $B'\in\mathcal B$ such that $B'\equiv A_y$, consequently $y<x$, which is impossible.
 A: Your proof is overly complicated, and the reliance on the axiom of choice is itself a bad direction. The construction ultimately leads to Hartogs' theorem which implies the axiom of choice follows from the comparability of any two cardinals. If we had used the axiom of choice to prove Hartogs' theorem, then this would be circular.
The proof is simpler and tantamount to just verifying by hand. Given a non-empty family $\cal B\subseteq A$, pick some $A$ such that $[(A,\leq)]$ is in $\cal B$. If this is a minimal element, we're done. So let's assume it's not.
Now each $[(B,\leq_B)]\in\cal B$ such that $[(B,\leq_B)]\prec[(A,\leq)]$ embeds into a unique proper initial segment of $A$. So each such $B$ gives us a unique $x$ such that $(B,\leq_B)$ is isomorphic to $A_x=\{a\in A\mid a<x\}$. Moreover, since $A$ is not the minimal element of $\cal B$, the set of $\{x\in A\mid\exists[(B,\leq_B)]\in\mathcal B, B\equiv A_x\}$ is non-empty. Since $\leq$ is a well-ordering of $A$, there is a minimal $x$.
Now show that $B$ which corresponds to this minimal $x$ is the minimal element of $\cal B$.
