Ordinal Subset of Ordinal is Initial Segment

Theorem
Let $$S$$ 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
We remind ourselves that the Ordering on an Ordinal is Subset Relation.

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

Then it follows, by the definition of minimal, that $$\forall x \in S: x \subset a$$ and hence $$x \notin S - T$$, so $$x \in T$$.

Thus by definition of segment, $$S_a \subseteq T$$.

Now let $$b \in T$$.

Then by definition of ordinal, $$T_b = b = S_b$$.

Now, if $$a \subset b$$ then $$a \in S_b$$.

So $$a \in T_b$$ and hence $$a \in T$$.

But $$a \in S - T$$, so by definition of set difference, this is not the case.

So $$a \not \subset b$$.

We have that an ordinal is well-ordered by $\subseteq$.

So by the definition of well-ordering, $$\subseteq$$ is a total ordering on $$S$$

So it follows that $$b \subseteq a$$.

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

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

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

Thus $$T = S_a$$.