Subset of Well-Ordered Set is Well-Ordered/Proof 1

Proof
First suppose that $T = \O$.

From Empty Set is Subset of All Sets, $T$ is a subset of $S$.

By Empty Set is Well-Ordered, $\struct {\O, \preceq'}$ is a well-ordered set.

Otherwise, let $T$ be non-empty.

Let $X \subseteq T$ such that $X \ne \O$ be arbitrary.

Such a subset exists, as from Set is Subset of Itself, $T$ itself is a subset of $T$.

By Subset Relation is Transitive, $X \subseteq S$.

By the definition of a well-ordered set, $X$ has a smallest element under $\preceq$.

That is:
 * $\forall y \in S: x \preceq y$

Hence as $T \subseteq S$:
 * $\forall y \in T: x \preceq y$

Because $\preceq'$ is the restriction of $\preceq$ to $T$:
 * $\forall y \in T: x \preceq' y$

and so $x$ is the smallest element of $X$ under $\preceq'$.

It follows by definition that $\struct {T, \preceq'}$ is a well-ordered set.