Is a closure-preserving union of closed paracompact subspaces paracompact? Let $X$ be a space. Suppose that $X$ has a closure-preserving closed covering $\mathcal{A}$ by paracompact subspaces. Then $X$ is regular.

Does it follow that $X$ itself is paracompact?

The answer is affirmative under the stronger assumption that $\mathcal{A}$ is hereditarily closure-preserving, and in particular under the assumption that $\mathcal{A}$ is locally-finite.
Note that any point-finite closure-preserving closed family is necessarily locally-finite, so perhaps I'm just not looking hard enough for a counterexample.

If the previous question is answered in the negative, then is there a counterexample where $\mathcal{A}$ is point-countable?

Edit: Thanks to Brian Scott for pointing out that my own attempt at a counterexample was flawed.
 A: This is a minor modification of the nice example due to H.B. Potoczny in the paper noted by Ulli in the comments.
Let $X=\omega_1\times\omega_1$, and let $D=\{\langle\alpha,\alpha\rangle:\alpha\in\omega_1\}$ be the diagonal. For convenience let $p_\alpha=\langle\alpha,\alpha\rangle$ for each $\alpha\in\omega_1$. Points of $X\setminus D$ are isolated. For each $\alpha\in\omega_1$ let
$$U_\alpha=\big(\{\alpha\}\times\omega_1\big)\cup\big(\omega_1\times\{\alpha\}\big)\,,$$
and for each finite $F\subseteq U_\alpha\setminus\{p_\alpha\}$ let $U_\alpha(F)=U_\alpha\setminus F$; these sets $U_\alpha(F)$ are the basic open nbhds of $p_\alpha$. It’s straightforward to verify that this defines a clopen base $\mathscr{B}$ for a Hausdorff topology, so $X$ is a Tikhonov space.
However, $X$ is not normal, so it is not paracompact. Let $H=\{p_n:n\in\omega\}$ and $K=D\setminus H$; $H$ and $K$ are disjoint closed sets. Let $V$ be any open nbhd of $H$; for each $n\in\omega$ there is a $U_n(F_n)\in\mathscr{B}$ such that $p_n\in U_n(F_n)\subseteq V$. Let $C=\bigcup_{n\in\omega}F_n$; $C$ is a countable subset of $X$, so there is an $\alpha\in\omega_1$ such that $\alpha\ge\omega$ and $\xi,\eta<\alpha$ for each $\langle\xi,\eta\rangle\in C$. Then $p_\alpha\in K$, since $\alpha\ge\omega$, and I claim that $p_\alpha\in\operatorname{cl}V$, so that $K$ cannot have an open nbhd disjoint from $V$. To see this, let $U_\alpha(F)\in\mathscr{B}$ be any basic open nbhd of $p_\alpha$; then
$$\langle n,\alpha\rangle,\langle\alpha,n\rangle\in U_\alpha(F)\cap U_n(F_n)\subseteq U_\alpha(F)\cap V$$
for all but finitely many $n\in\omega$, since $F$ is finite.
Finally, if $\alpha<\beta<\omega_1$ let
$$K(\alpha,\beta)=\{p_\alpha,p_\beta,\langle\alpha,\beta\rangle,\langle\beta,\alpha\rangle\}\,,$$
and let $\mathscr{K}=\{K(\alpha,\beta):\alpha<\beta<\omega_1\}$; this is clearly a cover of $X$ by compact sets and hence by paracompact closed sets, and I claim that it is closure-preserving. Let $\mathscr{A}\subseteq\mathscr{K}$, and let $A=\bigcup\mathscr{A}$. If $p_\gamma\in\operatorname{cl}A$ for some $\gamma\in\omega_1$, then there is a $K(\alpha,\beta)\in\mathscr{A}$ such that $U_\gamma\cap K(\alpha,\beta)\ne\varnothing$, but then $\gamma\in\{\alpha,\beta\}$, and $p_\gamma\in K(\alpha,\beta)\subseteq A$. Clearly the only isolated points in $\operatorname{cl}A$ are those already in $A$, so $A$ is closed, and $\mathscr{K}$ is closure-preserving.
The family $\mathscr{K}$ is not, however, point-countable: for each $\alpha<\omega_1$, $p_\alpha\in K(\alpha,\beta)$ for each $\beta$ such that $\alpha<\beta<\omega_1$.
