From Jech's Set Theory:
Lemma 2.11. (iii) If $α\ne β$ are ordinals and $α ⊂ β$, then $α ∈ β$.
Proof: If $α ⊂ β$, let $γ$ be the least element of the set $β − α$. Since $α$ is transitive, it follows that $α$ is the initial segment of $β$ given by $γ$. Thus $α = \{ξ ∈ β : ξ<γ\} = γ$, and so $α ∈ β$.
As $\alpha$ is transitive, we have $x\in\alpha\implies x\subset\alpha$, but I can't understand how the equality
$$α = \{ξ ∈ β : ξ<γ\}$$
is reached.
Since $\gamma$ is the least element of $\beta-\alpha$, we have that if $x\in\alpha$, then $x\lt \gamma$. Thus $\alpha\subseteq \{\zeta\in\beta\mid \zeta\lt\gamma\}$. Conversely if $\zeta\in\beta$ has $\zeta\lt\gamma$, then the minimality of $\gamma$ means that $\zeta\in\alpha$. Thus, $\{\zeta\in\beta\mid \zeta\lt\gamma\}\subseteq\alpha$, giving equality.