A set $S$ is said to be transitive if $x ⊆ S$ for every $x ∈ S$.
A total order $≤$ on a set $S$ is said to be a well-ordering if for every non-empty subset $A ⊆ S$, there exists an element $m ∈ A$ such that $m ≤ x$ $∀x ∈ A.$
An ordinal number is a set that is transitive and is well-ordered by the relation $α ≤ β ⇔ α ∈ β$ or $α = β$ or equivalently $α < β ⇔ α ∈ β$.
Now given that $α$ is an ordinal number and $β ⊆ α$ has the property $(∀x < y)(x ∈ β) ⇒ y ∈ β$ for every $y ∈ α$. I need to prove that $β = α$.
Here is my proof:
Suppose, in order to get a contradiction, that $\beta\ne \alpha$ .Then as $\beta\subseteq\alpha$, we have that $\alpha-\beta\ne \emptyset$ and hence $\alpha - \beta$ is a non-empty subset of the ordinal number $\alpha$. By well-orderedness of the ordinal $\alpha$, there exists an element $x\in \alpha-\beta$ such that $\forall y\in \alpha-\beta, x\le y$
Now consider any $\lambda\in \alpha$ such that $\lambda<x$ (Here is a possible problem which I have specified below). As if $\lambda\in \alpha-\beta$, then $x$ is not the least element of $\alpha-\beta$ which contradicts to the fact that $\forall y\in \alpha-\beta, x\le y$, there must be $\lambda\notin \alpha-\beta$ and hence $\lambda\in \beta$.
But then as we have $\lambda<x$ and $\lambda\in \beta$, by the property, $x\in\beta$, contradicts the fact that $x\in \alpha-\beta$.
Thus it cannot be that $\beta\ne \alpha$. Thus...
I have come up with the above proof but I realised that it might be not possible that there exists $\lambda\in \alpha$ such that $\lambda<x$.
$x=\emptyset\iff\emptyset\not\in\beta$. This is because $x$ is the least element of $\alpha$ that isn't in $\beta$, and $\emptyset$ is the least element of $\alpha$. However, if the property is true of $\beta$, then $\emptyset$ is in $\beta$ because the antecedent is vacuously true for $y=\emptyset$. Therefore the property itself asserts that $\emptyset\in\beta$.
Therefore we know that $\emptyset\in\beta$ and $x\neq\emptyset$. Therefore $x$ has elements, and so there is a $\lambda<x$.