Finite Totally Ordered Set is Well-Ordered

Theorem
Every finite totally ordered set is well-ordered.

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

From Condition for Well-Foundedness, $\left({S, \preceq}\right)$ is well-founded iff there is no infinite sequence $\left \langle {a_n}\right \rangle$ of elements of $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$.

If it were the case that $\left({S, \preceq}\right)$ had such an infinite sequence, then at least some of the elements would be repeated in that sequence.

So there would be, for example:
 * $s_i \preceq s_j \preceq s_k \preceq s_i$

and $\preceq$ would therefore not be transitive and so not a totally ordered set.

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

The result follows from the definition of well-ordered set.