Let $M\models T$ and $C\subset B\subset M$ be finite subsets. Let $a\in acl(C)$. How can we show that $tp(a/B)$ does not divide over $C$.
Here is my attempt: Suppose $tp(a/B)$ divides over $C$. Then there is a $C$-indiscernible sequence $(\bar{b}_i)_{i<\omega}$ such that $\bar{b}_0=\bar{b}$ where $\bar{b}$ is an enumeration of $B$, and $\{\phi(x,\bar{b}_i) : i<\omega\}$ is $k$-inconsistent for some $k<\omega$. Since $a\in acl(C)$, there is finitely many copies of $a$ over $C$ which is a contradiction !!!
Yes, this is correct. Let's make that last sentence precise, for completeness sake.
Let $\{a_1, \ldots, a_n\}$ be the conjugates of $a$ over $C$, with $a_1 = a$. Since $\models \phi(a, \bar{b})$ we have that for every $i < \omega$ there is $1 \leq j_i \leq n$ such that $\models \phi(a_{j_i}, \bar{b}_i)$. So by the pigeonhole principle there are $1 \leq j \leq n$ and an infinite $I \subseteq \omega$ such that $\phi(a_j, \bar{b}_i)$ for all $i \in I$. This contradicts $k$-inconsistency.