Enumerate $\mathcal{A}$ as $\langle A_\alpha:\alpha\in\omega_1\rangle$. Enumerate $[\omega_1]^{<\omega}\times\omega$ as $\bigl<\langle F_\alpha,n_\alpha\rangle:\alpha\in\omega_1\bigr>$, with $F_\alpha\subseteq\alpha$ always. Define recursively a sequence of neighourhoods: $U_\alpha$ of $A_\alpha$ such that $U_\alpha\cap(\bigcup\{A_\beta:\beta\in F_\alpha\}\cup n_\alpha)=\emptyset$. Then $\mathcal{U}=\{U_\alpha:\alpha\in\omega_1\}\cup\bigl\{\{n\}:n\in\omega\bigr\}$ is an open cover. If $\mathcal{V}$ is a finite subfamily of $\mathcal{U}$ then there is an $\alpha$ such that $\mathcal{V}\subseteq\{U_\beta:\beta\in F_\alpha\}\cup\bigl\{\{i\}:i\in n_\alpha\bigr\}$. Then $U_\alpha\cap\bigcup\mathcal{V}=\emptyset$, so the point $A_\alpha$ is not in $\operatorname{St}(\bigcup\mathcal{V},\mathcal{U})$.
Addendum (2022-01-30): this proof works for every (M)AD family of any cardinality; the conclusion is that $\Psi(\mathcal{A})$ is never starcompact.