In a stable theory, if we have a finite tuple $a$, and $A\subset$acl($a$). Can we find a finite subset of $A$, say $F$, such that $A\subset$acl($F$)?
I find that it is exactly Hilbert's basis theorem in the theory of Rings, and it is implied by some kind of descending properties. So I believe it's still true in a stable theory. But yet I haven't prove it.
I'm wondering if there is a counterexample or how to prove it.
Consider the theory with sorts $P$ and $(S_i)_{i\in\omega}$ and unary function symbols $\pi_i\colon P\to S_i$ for all $i$. Let $T$ be the complete theory of the structure $M$ in which each $S_i$ is a countably infinite set, $P$ is the product of these sets, and each $\pi_i$ is the projection onto the $S_i$ factor.
Let $a$ be an element of the $P$ sort, and let $A=\{\pi_i(a)\mid i\in\omega\}$. Then $A$ is a subset of $\mathrm{acl}(a)$, but there is no finite subset $F$ of $A$ such that $A\subseteq \mathrm{acl}(F)$. Indeed, the elements of $F$ live in finitely many of the sorts $S_i$, and there are automorphisms of $M$ fixing these elements and arbitrarily permuting the other sorts.
I must say I don't exactly see the connection to the Hilbert basis theorem - this theorem is about finitely generated rings and the "ideal generated by" operation, and I think the model theoretic acl operation is not a perfect analogy for either the "ring generated by" or "ideal generated by" operations.
I think a better algebraic analogue of your question is this fact: if an algebraically closed field has finite transcendence degree over the prime field, then any subfield will also have finite transcendence degree. And this suggests the following resolution: your question has a positive answer in any theory where acl satisfies exchange (i.e., where acl is a pregeometry). In fact, in this case the set $F$ can always be chosen to have size at most the length of the finite tuple $a$.