Using the Subset Axioms to prove the existence of a set I'm going through Enderton's Elements of Set Theory, as I heard it is a gentle introduction to set theory. I'm a little confused on how the subset axioms are used. In the text, the axiom is given as:

Subset Axioms For each formula $\underline{\hspace{1cm}}$ not containing $B$, the following is an axiom:
  $$
\forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge \underline{\hspace{1cm}}).
$$

Later, the text asserts the existence of the set $\{A\cup X\ | \ X\in\mathscr{B}\}$, call it $\mathscr{D}$, saying that since $A\cup X\subseteq A\cup\bigcup\mathscr{B}$, $\mathscr{D}$ is a subset of $\mathscr{P}(A\cup\bigcup\mathscr{B})$. A subset axiom produces 
$$
\{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}.
$$
My question is, what exactly is that subset axiom? Would it be something of the form
$$
\forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [(x=t_1\cup t_2)\vee(x=t_1\cup t_3)\vee\cdots\vee(x=t_1\cup t_k)])
$$
where we take $A$ to be a particular instance of $t_1$, $\mathscr{P}(A\cup\bigcup\mathscr{B})$ to be a particular instance of $c$, and $t_2,\dots,t_k$ to be particular instances of members of $\mathscr{B}$?
Or perhaps something like
$$
\forall t_1\forall t_2\forall t_3\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [\exists t_3\in t_2(x=t_1\cup t_3)])
$$
where in this case $\mathscr{B}$ is a particular instance of $t_2$?, and then $\mathscr{D}$ would be the set $B$ which has been proven to exist. Thanks, right now I'm just not used to what should go in that blank spot for the formula.
 A: If you want to form the set 
$$
\{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}.
$$
then a natural instance of the axiom of separation (the "subset axiom") is
$$
(\exists B)(\forall t) ( t \in B \Leftrightarrow t\in\mathscr{P}(A\cup\bigcup\mathscr{B}) \land (\exists X)[X \in \mathscr{B} \land (\forall s) (s \in t \Leftrightarrow s \in A \lor s \in X)]).
$$
This formula is obtained by simply unpacking the original abbreviated formula, for example by expanding the "=" using its set-theoretic definition. 
There is a general principle that if you have assumed a formula of the form $(\forall z)\phi(z)$ then, for any set $Y$, the formula $\phi(Y)$ will be true as well. This is called (universal) instantiation. The way you get to the separation axiom above is to start with 
$$
(\forall C_1)(\forall C_2)(\forall C)(\exists B)(\forall t) ( t \in B \Leftrightarrow t\in C \land (\exists X)[X \in C_1 \land (\forall s) (s \in t \Leftrightarrow s \in C_2 \lor s \in X)]).
$$
and then instantiate it with $C_1 = \mathscr{B}$, $C_2 = A$, $C = \mathscr{P}(A\cup\bigcup \mathscr{B}) $. 
