Why does any transitive model satisfy extensionality?

I see it stated as something very clear, but i can't figure it out. i found a proof in Jech(old version) which goes through the concept of restricted formulas, which i don't quite understand. (i'm not sure i understand the notation in his Lemma 10.1, the statement numbered 10.5) Is there a way to show it directly?

Let $\mathbf{M}$ be transitive, and suppose $x , y \in \mathbf{M}$ are distinct. Then, by extensionality in $V$, there must be (in $V$) a $u$ which belongs to either $x \setminus y$ or $y \setminus x$. In either case the transitivity of $\mathbf{M}$ implies that $u \in \mathbf{M}$, and so $\mathbf{M} \models \neg ( u \in x \leftrightarrow u \in y )$, or, more meaningfully $\mathbf{M} \models \neg ( \forall u ) ( u \in x \leftrightarrow u \in y )$. Therefore $$\mathbf{M} \models ( \forall x ) ( \forall y ) ( x \neq y \rightarrow \neg ( \forall u ) ( u \in x \leftrightarrow u \in y ))$$ which (logically equivalent to) the Axiom of Extensionality in Jech's text.

The $\in$-relation of transitive models is the true $\in$ of the universe, and we know that the universe satisfies extensionality.

Now suppose $M$ is a transitive model, and that $x,y\in M$. By transitivity we know that $x,y\subseteq M$ as well. Therefore $M$ knows all of their elements. If $x\neq y$ then without loss of generality there is some $z\in x\setminus y$, but $M$ knows about $z$ and since $M$ agrees with the universe about $\in$, $M$ also knows that $z\in x\setminus y$.

So whenever $M$ is transitive and $x,y\in M$ are distinct elements, $M\models x\neq y$.