From $\textit{The Higher Infinite}$ by Kanamori, in his proof (page 39 - 41) that $\kappa$ is weakly compact if and only if $\kappa$ satisfies the Extension Property, the proof does something that I am not entirely sure is necessary:
The relevant portion of this proof is as follows:
Let $\Sigma$ be a $\kappa$ satisfiable set of $L_{\kappa,\kappa}$ sentences. $R_1$ is the definable satisfaction relation for set models of $V_\kappa$. $R_2$ is another relation (but that is not relevant at this point in the proof). He then manages to prove that there is a transitive set $X$ properly containing $V_\kappa$ such that
$\langle V_\kappa, \in, R_1, R_2 \rangle \prec \langle X, \in, S_1, S_2 \rangle$
$\langle X, \in, S_1, S_2) \models \Sigma \text{ has a model }$
I think that the proof should be finished at this point: Let $\mathcal{M}$ in $X$ be such a model. By the absoluteness of the satisfaction relation $S_1$, $\mathcal{M}$ is a model of $\Sigma$ in $V$. This should conclude the proof.
However, Kanamori continues. Since $\kappa$ is inaccessible, $V_\kappa \models ZFC$. By elementarily, $X \models ZFC$. Being inaccessible is absolute for transitive model of $ZFC$, so $X \models \kappa$ is inaccessible. Then he applies the Lowenheim Skolem Theorem in $X$ for $L_{\lambda, \lambda}$ for inaccessible cardinal $\lambda$ to prove that (when $\lambda = \kappa$)
$\langle X, \in, S_1, S_2 \rangle \models \Sigma$ has a model $\mathcal{M}$ with domain $\subseteq \kappa$
Then he applies the absoluteness of the satisfaction relation $S_1$, to claim that $\mathcal{M}$ is really a model of $\Sigma$.
My question is why does he have to produce a model in $X$ of $\Sigma$ with domain a subset of $\kappa$ to apply the absoluteness argument of the satisfaction relation. Would it not be enough just to produce any model in $X$. Thanks for any clarification