From this wikipedia article:
For each first-order $\sigma $-formula $\varphi(y,x_{1}, \ldots, > x_{n}) \,,$ the axiom of choice implies the existence of a function :$f_{\varphi}: M^n\to M$
such that, for all $a_{1}, \ldots, a_{n} \in M$, either :$M\models\varphi(f_{\varphi} (a_1, \dots, a_n), a_1, \dots, a_n)$
or :$M\models\neg\exists y\, \varphi(y, a_1, \dots, a_n) \,.$
Applying the axiom of choice again we get a function from the first-order formulas $\varphi$ to such functions $f_{\varphi} \,.$
The family of functions $f_{\varphi}$ gives rise to a preclosure operator $F $ on the power set of $M $ :$F(A) = \{f_{\varphi}(a_1, > \dots, a_n) \in M \mid \varphi \in \sigma ; \, a_1, \dots, a_n \in A \} $
for $A \subseteq M \,.$
Iterating $F $ countably many times results in a closure operator $F^{\omega} \,.$ Taking an arbitrary subset $A \subseteq M$ such that $\left\vert A \right\vert = \kappa$, and having defined $N = F^{\omega}(A) \,,$ one can see that also $\left\vert N \right\vert = \kappa \,.$
I can see that $\kappa \leq \left\vert N \right\vert$ on account of properties of closure operators, yet $\left\vert N \right\vert \leq \kappa$ seems less obvious to me.
It seems to me each $f_{\varphi}$ maps $A^{n_{\varphi}}$ to $M_{\varphi} \subseteq M$, where $\left\vert M_{\varphi} \right\vert \leq \left\vert A^{n_{\varphi}} \right\vert = \kappa^{n_{\varphi}}$.
Since we iterate until closure, this seems to give rise to $\left\vert N \right\vert \leq \sum_{\varphi \in \sigma} \kappa^{n_{\varphi}^{\omega}}$, where $n_{\varphi}$ is raised to $\omega$ or countable infinity.
I am not sure this is a correct approach, or how to avoid $\kappa < \left\vert N \right\vert$.
Edit: It does say later in the proof that $\left\vert F(A) \right\vert \leq \left\vert A \right\vert + \left\vert \sigma \right\vert + \aleph_0$. However, I can't see how that can be guaranteed.