Strictly Well-Founded Relation determines Strictly Minimal Elements

Theorem
Let $A$ be a class.

Let $\prec$ be a foundational relation on $A$.

Let $B$ be a nonempty class and suppose that $B \subseteq A$.

Then $B$ has a $\prec$-minimal element.

Proof
Let $\prec' = \left({B \times B}\right) \cap \prec$.

$\prec'$ is a foundational relation by Foundational Relation Subset.

By Well-Founded Relation Determines Minimal Elements/Lemma, $B$ has a $\prec'$-minimal element, $m$.

Suppose that for some $x \in B$, $x \prec m$.

Then $\left({x,m}\right) \in \prec$.

Since $x,m \in B$, $\left({x,m}\right) \in B \times B$.

By the definition of $\prec'$, $\left({x,m}\right) \in \prec'$, so
 * $x \prec' m$.

Since $m$ is $\prec'$-minimal in $B$, $x = m$.

As we have shown that $\forall x\in B: x \prec m \implies x = m$, $m$ is $\prec$-minimal in $B$.

Also See
and weaker results that don't require the Axiom of Foundation.
 * Well-Founded Relation with Small Initial Segments Determines Minimal Elements‎
 * Well-Ordering Determines Minimal Elements,