This is another exercise from Kees Doets Basic Model Theory. Here's the idea. It's well known that the downward Löwenheim-Skolem theorem follows as an easy corollary of the following lemma using Skolem functions:
Suppose that $\mathcal{B}$ is an $L$-model, that $X \subset B$ and that $\aleph_0, |L|, |X| \leq \mu \leq |B|$. Then a submodel $\mathcal{A} \subset \mathcal{B}$ exists such that $X \subset A$ and $|A| = \mu$.
So the idea is to prove the above lemma. There are several ways of doing this, but one suggestion by Doets that I found interesting is to use monotone operators. For starters, set $A_0$ as a set of power $\mu$ such that $X \subset A_0$ and consider an operator $\gamma: \mathcal{P}(B) \to \mathcal{P}(B)$ such that
$\gamma(Y) = Y \cup A_0 \cup \{\text{all constants from }\mathcal{B}\} \cup \{\text{all values of functions from } \mathcal{B} \text{ with arguments from }Y\}$.
Note that $\gamma$ is obviously monotonic, so it has a least fixed point, i.e. a set $Y$ such that $\gamma(Y) = Y$. Let $A$ be such least fixed point. Doets suggests that it is the desired submodel. Now, it's obvious that $A \subset B$ and that it's closed under functions and constants from $\mathcal{B}$. My only worry is with the cardinality of $A$; how do I know $|A| = \mu$?
It seems clear enough that $\mu \leq |A|$, since $A_0 \subseteq A$. It's also clear that the cardinality of the set of constants and function values from $\mathcal{B}$ has cardinality at most $\mu$, since $|L| \leq \mu$. But it's not clear to me that this enough to conclude that $|A| = \mu$, since I don't know the closure ordinal for $\gamma$. Any ideas on how to get the result?
All of the functions involved in forming $\gamma(Y)$ are finitary, so $\gamma$ satisfies $\gamma(\bigcup A_n)=\bigcup \gamma(A_n)$ whenever $A_1\subseteq A_2\subseteq\dots$ is a nested sequence. It follows that you only need to iterate it $\omega$ times to get a fixed point.