 Solution

Put $B=N\setminus \mathrm{acl}(A)$. 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 $C\cap \mathrm{acl}(A)\ne \varnothing$, 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 $\mathrm{acl}(A)\subseteq M\cap N$, while omission of $B=N\setminus \mathrm{acl}(A)$ gives the reverse inclusion. Therefore