Is x ⊆ S a well formed statement, when x is not a set? Let x be a object that is not a set
Let S be a set
Would the following statement:
x ⊆ S
evaluate to False, or considered not a well formed statement (as x is not even a set).
 A: In Zermelo-Fraenkel set theory, i.e. the standard axioms of set theory, every "object" is a set. So a statement like $x\subseteq S$ will always make sense, and have the same value of truth as
$$
\forall z:z\in x\implies z\in S
$$
"Everything is a set" can be confusing. "What are numbers then?" someone may ask. For example, natural numbers can be constructed as sets as
$0:=\emptyset$, $1:=\{0\}=\{\emptyset\}$, $2:=\{0,1\}=\{\emptyset,\{\emptyset\}\}$, $3:=\{0,1,2\}=\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$ and so on...
There other possible foundations for mathematics, like type theory, where there can be things that are not sets, so your question makes sense in that setting (but I do not know type theory).
A: In the standard Axiomatic Set Theory of Zermelo-Fraenkel all objects (i.e., variables that you can quantify or constants you can use in formulas) are sets, so, in that case, a formula $x\subseteq S$ where $x$ is not a set makes little to no sense. 
What some authors may do is extend the language to include symbols for classes (as done in Introduction to Axiomatic Set Theory by  Takeuti, G. and Zaring, W.M.), what must be clear is that you can't use any quantifier at those symbols, they represent collections of sets and are not objects in the strict sense, so, supposing that $x$ is a class but not a set, $x\subseteq S$ makes sense but would always be a false formula if $S$ is a set, because a set cannot have a proper class as a "subset" (or "subclass"), you can prove that aplying the Axiom of Separation to $S$ and having that $x$ is a set, thus a contradiction.
