Here's an exercise from the really nice Hinman's book "Foundamentals of Mathematical Logic". The minimal structures are those whose definable sets are finite or cofinite. The algebraic closure (acl) of $X$ is a set with elements from some finite set that is definable with parameters from $X$. I've stuck with the last sentence from the hint. I've proved the union contains at most $nm$ elements but can't see how to get a contradiction from it. Will be greatful for all your tips.
P.S. $\exists^{=k}xF$ means there are exactly $k$ values of $x$ s.t. $F$ is true.
Since $\mathfrak{A}$ is infinite, because $\bigcup_{i\leqslant m}\neg\psi_{a'_i,c}^\mathfrak{A}$ is finite its complement in $\mathfrak{A}$ must be infinite. Let $b'\in\mathfrak{A}\setminus\bigcup_{i\leqslant m}\neg\psi_{a'_i,c}^\mathfrak{A}$ be arbitrary; then $b'$ satisfies $\psi(a'_i,y,c)$ for each $i\leqslant m$. Recall that $$\psi(a'_i,y,c)\equiv\phi(a'_i,y,c)\wedge\exists^{=m}x\phi(x,y,c).$$ In particular, $\phi(a'_i,b',c)$ holds for each $i\in\{0,\dots,m\}$, and so $\phi_{b',c}^\mathfrak{A}$ has size at least $m+1$. On the other hand, $\exists^{=m}x\phi(x,b',c)$ holds as well, so this gives the desired contradiction.