Finite Totally Ordered Set is Well-Ordered

Theorem
Every finite totally ordered set is well-ordered.

Proof
In the following, we suppose that $$\left({S_n, \preceq}\right) = \left\{{s_1, s_2, \ldots, s_n}\right\}$$ is a totally ordered set with $$n$$ elements, such that:
 * $$\forall s_i, s_j \in S_n: s_i \preceq s_j \iff i \le j$$

That is, the elements of $$S_n$$ are ordered by their indices.

Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition:
 * $$S_n$$ is well-ordered.

$$P(1)$$ just says $$S_1 = \left\{{s_1}\right\}$$ is well-ordered, which is trivially true.

This is our basis for the induction.

Basis for the Induction
$$P(2)$$ is the case $$S_2 = \left\{{s_1, s_2}\right\}$$.

We note that $$\left\{{s_1}\right\} \subseteq S_2$$ and $$\left\{{s_2}\right\} \subseteq S_2$$.

We also note that $$\forall x \in S_2: s_1 \preceq x$$, and so $$x$$ is the minimal element of $$S_2$$.

Both of these (trivially) have a minimal element and therefore $$S_2$$ is well-ordered.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:
 * $$\forall j \in \N: 1 \le j \le k: S_k$$ is well-ordered.

Then we need to show:
 * $$\forall j \in \N: 1 \le j \le k+1: S_{k+1}$$ is well-ordered.

Induction Step
This is our induction step:

Consider the totally ordered set $$S_{k+1}$$ with $$k+1$$ elements.

From Totally Ordered Subset, any subset of $$S_{k+1}$$ is also totally ordered.

But from the induction hypothesis, any such subset is well-ordered.

The set $$S_{k+1}$$ itself has a minimal element, that is, $$S_1$$.

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $$\forall n \in \N^*: S_n$$ is well-ordered.