Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2

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

Then:
 * $\preceq$ is a total ordering
 * $\left({S, \preceq}\right)$ has a smallest element.

Proof
By definition of well-ordered, every subset of $S$, including $S$ itself, has a smallest element under $\preceq$.

Now consider $X = \left\{{a, b}\right\}$ where $a, b \in S$.

Because $S$ is well-ordered under $\preceq$, $X$ has a smallest element.

So either $\min X = a$ or $\min X = b$.

If $\min X = a$, then $a \preceq b$.

If $\min X = b$, then $b \preceq a$.

So either $a \preceq b$ or $b \preceq a$.

That is, $a$ and $b$ are comparable.

As this applies to all $a, b \in S$, the ordering $\preceq$ is total.