Transitive Set is Proper Subset of Ordinal iff Element of Ordinal

Theorem
Let $A$ be an ordinal.

Let $B$ be any transitive set $B$.

Then:
 * $B \subsetneq A \iff B \in A$

Necessary Condition
Suppose that $B \in A$.

Because an ordinal is transitive, it follows that $B \subseteq A$.

Also, $B \ne A$ because the strict well-ordering on $A$ is, by definition, given by the $\in$-relation.

Therefore, $B \subsetneq A$, as desired.

Sufficient Condition
Suppose that $B \subset A$.

By the definition of set equality, the set difference $A \setminus B$ is non-empty.

By the definition of a well-ordering, there exists a minimal element $x$ of $A \setminus B$.

Since the strict well-ordering on $A$ is given by the $\in$-relation, it follows by the definition of a minimal element that:
 * $\forall y \in A \setminus B: y \notin x$

Therefore:

Suppose that $z \in B$.

Since $B \subset A$, it follows that $z \in A$.

Recall that the strict well-ordering on $A$ is given by the $\in$-relation.

From Well-Ordering is Total Ordering, it follows that $z \in x$ or $x = z$ or $x \in z$.

If $x = z$ or $x \in z$, then it follows by the transitivity of $B$ that $x \in B$.

This contradicts the definition of $x$.

Hence, $z \in x$.

That is, $B \subseteq x$.

We have shown that $x \subseteq B$ and $B \subseteq x$.

By Equality of Sets, it follows that $B = x \in A$, as desired.

Corollary
For ordinals $A$ and $B$, $A \subset B \iff A \in B$. This follows immediately from the fact that all ordinals are transitive.

Also see

 * Ordering on an Ordinal is Subset Relation
 * Initial Segment of Ordinal is Ordinal

Source

 * : $\S 7.7$, $\S 7.8$