Ordinal Subset of Ordinal is Initial Segment

Theorem
Let $$\left({S; \preceq}\right)$$ be an ordinal.

Let $$T \subset S$$ also be an ordinal.

Then $$\exists a \in S: T = S_a$$, where $$S_a$$ is the segment of $$S$$ determined by $$a$$.

Proof
Let $$a$$ be the minimal element of $$S - T$$.

Thus $$S_a \subseteq T$$.

Now let $$b \in T$$.

Then $$T_b = b = S_b$$.

Now, if $$a \prec b$$ then $$a \in S_b$$. So $$a \in T_b$$ and hence $$a \in T$$ which is not the case.

So $$b \preceq a$$.

But $$b \ne a$$ since $$b \in T$$.

Hence $$b \prec a$$, and so $$b \in S_a$$.

This proves that $$T \subseteq S_a$$.

Thus $$T = S_a$$.