Infinite Sequence Property of Well-Founded Relation/Reverse Implication/Proof 2

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.

Let $T \subseteq S$ have no minimal element.

Let $a_0 \in T$.

Since $a_0$ is not minimal in $T$, so it is possible to find $a_1 \in T: a_1 \prec a_0$.

Similarly, $a_1$ is not minimal in $T$, so it is possible to find $a_2 \in T: a_2 \prec a_1$.

Since each $a_{k+1}$ is chosen as an arbitrary element based on $a_k$, we cannot do this an infinite number of times using the usual principle of recursive definition. Rather, we must invoke the Axiom of Dependent Choice as follows:

Suppose that $a_k \in T$.

Then as $a_k$ is not minimal in $T$, we can find $a_{k+1} \in T: a_{k+1} \prec a_k$.

So, by the Axiom of Dependent Choice, it follows that $\forall n \in \N: \exists a_n \in T: a_{n+1} \prec a_n$.

Thus we have been able to construct an infinite sequence $\left \langle {a_n}\right \rangle$ in $T$ such that $\forall n \in \N: a_{n+1} \prec a_n$.