Löwenheim–Skolem theorem without parameters?

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On page 160 of Smullyan's "A Beginner's Guide to Mathematical Logic", the Löwenheim–Skolem theorem and the compactness theorem is stated and proved in the following manner:

Theorem L.S.C. If $S$ is a set of closed formulas without parameters, and all finite subsets of $S$ are satisfiable, then the entire set $S$ is satisfiable in a denumerable domain.

Proof. We are given a denumerable set $S$ of closed formulas without parameters such that all finite subsets of $S$ are satisfiable. As in the case of Propositional Logic, we arrange the elements of $S$ in some denumerable sequence $X_1,X_2,\ldots,X_n,\ldots$. We start the tableau with $X_1$ at the origin. This concludes the first stage, and at any stage $n$, we proceed systematically, as already explained, and also append $X_{n+1}$ at the end of every open branch. Since every finite subset of $S$ is satisfiable, then at no stage can the tableau close. Hence it has an infinite open branch $\theta$, and the set of formulas on $\theta$ is a Hintikka set (since the construction is systematic), and contains all the elements of $S$. Thus $S$ is satisfiable in the denumerable domain of the parameters.

I believe that this proof is quite straightforward, since constructing the tableau is very mechanical. What I don't understand here is: why is the theorem stated only for closed formulas without parameters? Does the same method of proof fails when some elements of $S$ has parameters in them (If this is the case, please enlighten me)? Why is it necessary to distinguish the L.S.C. theorem in two different versions - with/without parameters?

Thank you in advance.