Restriction of Total Ordering is Total Ordering

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

Let $T \subseteq S$.

Let $\preceq'$ be the restriction of $\preceq$ to $T \times T$.

Then $\preceq'$ is a total ordering of $T$.

Proof
By Restriction of Ordering is Ordering, $\preceq'$ is an ordering.

If $x, y \in T$, then since $T \subseteq S$: $x, y \in S$.

Since $\preceq$ is a total ordering:
 * $(x,y) \in \preceq$ or $(y,x) \in \preceq$.

Suppose that $(x,y) \in \preceq$.

Since $x,y \in T$: $(x,y) \in T \times T$.

Thus $(x,y) \in (T \times T) \cap \preceq = \preceq'$.

A similar argument shows that if $(y,x) \in \preceq$ then $(y,x) \in \preceq'$.

Thus $\preceq'$ is a total ordering of $T$.