= Solution
Put $B=N\setminus\operatorname{acl}(A)$. We seek a model containing $A$ and omitting every element of $B$. Every finite set of these omission requirements is satisfiable: if a finite $C\subseteq B$ met every model containing $A$, the supplied result would imply $C\cap\operatorname{acl}(A)\ne\varnothing$, a contradiction.
Apply the <compactness theorem> to the elementary-diagram formulation of these requirements, using the <Tarski-Vaught test> to axiomatize the selected elementary submodel. It gives a model $M$ containing $A$ and omitting all of $B$. Part (a) gives $\operatorname{acl}(A)\subseteq M\cap N$, while omission of $B=N\setminus\operatorname{acl}(A)$ gives the reverse inclusion. Therefore
$$
\boxed{M\cap N=\operatorname{acl}(A).}
$$
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