Questions about proof of consistency of GCH In Jech's Set Theory, Theorem 13.20 states the following:

If $V=L$ then $2^{\aleph_\alpha}=\aleph_{\alpha+1}$ for every $\alpha$.

In the proof, he writes the following:

Let $X\subseteq \omega_\alpha$. There exists a limit ordinal $\delta>\omega_\alpha$ such that $X\in L_\delta$. Let $M$ be an elementary submodel of $L_\delta$ such that $\omega_\alpha\subseteq M$ and $X\in M$, and that $|M|=\aleph_\alpha$.

How do we know that such an $M$ exists? So far as I understand, Lowenheim-Skolem gives an elementary submodel $M$ of $L_\alpha$ such that $|M|=\aleph_\alpha$, but I don't know how to guarantee $\omega_\alpha\subseteq M$ and $X\in M$.
He then writes

By the Condensation Lemma 13.17, the transitive collapse $N$ of $M$ is $L_\gamma$ for some $\gamma\leq \delta$. Clearly, $\gamma$ is a limit ordinal, and $\gamma<\omega_{\alpha+1}$ because $|N|=|\gamma|=\aleph_\alpha$.

Apparently it's meant to be clear, but I don't actually see why $\gamma$ has to be a limit ordinal (or why it's necessary). I'm also not quite sure why the chain of equalities as Jech writes it is true. I would believe $\aleph_\alpha=|M|\geq |N|=|L_\gamma|=|\gamma|$, but I don't see why equality has to hold (or why he wrote it in that order, but that's more a matter of style).
 A: Re: your first question, it sounds like you've seen the following version of the downward Lowenheim-Skolem theorem:

(dLS) Every structure $\mathcal{M}$ in a language $L$ of cardinality $\le\kappa$ (with $\kappa$ infinite) has an elementary substructure of size $\le\kappa$.

In fact, a seemingly-stronger result also holds:

(dLS') For every infinite $\kappa$, every structure $\mathcal{M}$ in a language of size $\le\kappa$, and every $A\subseteq\mathcal{M}$ with $\vert A\vert\le\kappa$, there is an elementary substructure of $\mathcal{M}$ containing $A$ with cardinality $\le\kappa$.

This is what Jech is using here (with $\kappa=\omega_\alpha$ and $A=\omega_\alpha\cup\{X\}$). However, dLS' is in fact an immediate corollary of dLS. This is a good exercise; here's a hint:

 Think about expanding the language of $\mathcal{M}$ by constant symbols naming the elements of $A$.


For your next two questions, I think the key thing you're missing is that the Mostowski collapse map is an isomorphism.
First, note that we have $$N\cong M\equiv L_\delta.$$ Since $\delta$ was a limit ordinal, $L_\delta\models$ "There is no greatest ordinal," so consequently we get $N\models$ "There is no greatest ordinal." If $N=L_{\gamma+1}$ then $N$ would think that there was a greatest ordinal (namely $\gamma$).
Finally, re: the chain of equalities, since $M\cong N$ we have $\vert M\vert=\vert N\vert$.
