Infinite Sequence Property of Well-Founded Relation/Reverse Implication/Proof 1

Theorem
Let $\left({S, \preceq}\right)$ be a ordered set.

Suppose that there is no infinite sequence $\left \langle {a_n}\right \rangle$ of elements of $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$.

Then $\left({S, \preceq}\right)$ is well-founded.

Proof
Suppose $\left({S, \preceq}\right)$ is not well-founded.

So by definition there exists a non-empty subset $T$ of $S$ which has no minimal element.

Let $a \in T$.

Since $a$ is not minimal in $T$, we can find $b \in T: b \prec a$.

Since this holds for all $a \in T$, $\prec \restriction_{T \times T}$, the restriction of $\prec$ to $T \times T$, is a right-total endorelation on $T$.

So, by the Axiom of Dependent Choice, it follows that there is an infinite sequence $\left \langle {a_n}\right \rangle$ in $T$ such that $\forall n \in \N: a_{n+1} \prec a_n$.