To prove that $S$ has $(C7)$ you need $S$ to satisfy three things,
- For all $c,d\in C$, if $(c=d)∈s∈S$ then $s∪\{d=c\}∈S$
This part is obvious from the definition of $S$, as if $(\mathfrak A, b_0,\ldots)$ witness $s\in S$, it will also witness $s∪\{d=c\}∈S$
- For all basic terms $t$ and for all $c\in C$, if $(t=c),φ(t)∈s\in S$, then $s\cup \{φ(c)\}∈S$
Same argument from above will show that that holds
- For all basic terms $t$, there exists $e\in C$ such that if $s\in S$ implies $s\cup\{t=e\}∈S$
To show that, first note that if $t$ is a basic term that is not of the form $d_r$ for some rational $r$ and $s∈S$ (where $s=s_0∪T\cup \cdots$) with witness $(\mathfrak A, b_0,\ldots)$ then $s_0$ doesn't contain any mention of $t$ and any mention of some $c\in C$ (as it contains only finitely many such $c$), so we can just let $(\mathfrak A', b_0,\ldots)$ be the same as $(\mathfrak A, b_0,\ldots)$ but where we assigne $c$ to be $t$.
So all we need to show is that if $d_r∈D$ and $s\in S$, then $s\cup\{d_r=e\}∈S$ for some $e\in C$. The claim Keisler proves is that if $e$ does not accure in $s_0$, then indeed $s\cup\{d_r=e\}∈S$, which proves that $S$ has $(C7)$