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 $(C7)$
To actually show the last paragram consider $s=s_0\cup T\cup \{d_{r_0}<d_{r_1}<d_{r_2}<\cdots<d_{r_n}\}∈S$ where $r_0<r_1<\cdots<r_n$ are rationals, for now assume $r$ appears in $\{d_{r_0}<d_{r_1}<d_{r_2}<\cdots<d_{r_n}\}$, so we call $r=r_p$ where $0\le p\le n$ and $d_r=d_{r_p}$.
Let $s_1(\bar c, e, d_{r_0},\cdots,d_{r_n})=s_0(\bar c, d_{r_0},\cdots,d_{r_n})\cup \{d_r=e\}$ where $e$ does not appear in $s_0$, and consider $s'=s_1\cup T\cup \{d_{r_0}<d_{r_1}<d_{r_2}<\cdots<d_{r_n}\}$, we want to show that $s'∈ S$.
Let $α<ω_1$ be arbitrary, we want to find a model $\mathfrak B_α$ in $K$ and assignments $b_0,\ldots,b_n$ to $d_{r_0},\ldots,d_{r_n}$ such that $(\mathfrak B_α,b_0,\ldots)⊨ ∃\bar x∃y (\bigwedge s_1(\bar x, y, d_{r_0},\ldots d_{r_n}))$ such that [the confusing condition on the $b_i$s is true].
Now if $\mathfrak A_α$ with $a_0,\ldots, a_n$ is witness for $(\mathfrak A_α,a_0,\ldots)⊨ ∃\bar x (\bigwedge s_0(\bar x, d_{r_0},\ldots d_{r_n}))$, then note that by letting $\mathfrak B_\alpha=\mathfrak A_α$, and $a_i=b_i$ we are finish (witness with $y$ being $b_p$)
All we left to do is to show why we can assume $r=r_p$ for some $p$.
Assume $r$ is not in there, so let $p$ be such that $r_p<r<r_{p+1}$ (the cases where $r<r_0$ or $r>r_n$ are pretty much identical so I won't talk about them).
As before we shall open the definition, let $α<ω_1$ be arbitrary, we want to find a model $\mathfrak B_α$ in $K$ and assignments $b_0,\ldots,b_n$ to $d_{r_0},\ldots,d_{r_n}$ as well as an assignment $b$ for $d_r$ such that $(\mathfrak B_α,b_0,\ldots,b_n,b)⊨ ∃\bar x(\bigwedge s_0(\bar x, d_{r_0},\ldots d_{r_n}))$ (note here that $s_0$ does not depends on $d_r$) such that $$α≤b_0, b_0+α≤b_1,\cdots, b_p+α≤b, b+α≤b_{p+1},\cdots b_{n-1}+α≤b_n$$.
For this, consider $β=α+α<ω_1$ and the model $\mathfrak A_{β}$ and the assignments $a_0,\ldots, a_n$ to $d_{r_0},\ldots,d_{r_n}$ such that $(\mathfrak A_α,a_0,\ldots,a_n)⊨ ∃\bar x(\bigwedge s_0(\bar x, d_{r_0},\ldots d_{r_n}))$ such that $$β≤a_0, a_0+β≤a_1,\cdots, a_p+β≤a_{p+1},\cdots a_{n-1}+β≤a_n$$ (Such $\mathfrak A_β, a_i$ exists because $s\in S$)
Note that we also have we also have $$α≤a_0, a_0+α≤a_1,\cdots, a_p+α≤a_p+α, (a_p+α)+α≤a_{p+1},\cdots a_{n-1}+α≤a_n$$
So we finish the proof by letting $\mathfrak B_α=\mathfrak A_β$, $b_i=a_i$ and $b=b_p+α$.