Proper Well-Ordering determines Smallest Elements
Let $S$ be a class.
Let $\preceq$ be a proper well-ordering on $S$.
Let $B$ be a nonempty subclass of $S$.
Then $B$ has a $\preceq$-smallest element.
Because $B \ne \varnothing$ it follows that $B$ has an element $x$.
If $x$ is the smallest element of $B$ then the theorem holds.
Otherwise, there is an element $y \in B$ such that $x \npreceq y$.
By Well-Ordering is Total Ordering, $y \prec x$.
Thus $S_x$, the $\preceq$-initial segment of $x$, is not empty.
By the hypothesis, $S_x$ is a set.
By Intersection is Subset, $B \cap S_x \subseteq S_x$
Thus by the subset axiom, $B \cap S_x$ is a set.
Thus by the definition of proper well-ordering, $B \cap S_x$ has a smallest element $a$.
Let $z$ be any element of $B$.
Suppose for the sake of contradiction that $z \prec a$.
Then by Extended Transitivity, $z \prec x$, so $z \in B \cap S_x$.
But this contradicts the fact that $a$ is the smallest element of $B \cap S_x$.
Thus $z \nprec a$.
Since $\preceq$ is a total ordering, $a \preceq z$.
As this holds for all $z \in B$, $a$ is the smallest element of $B$.