Is this space paracompact? Is this space $F[R]$ with the Pixley-Roy topology paracompact?
In general, when the space $F[X]$ is paracompact for general topological space?

Definition of Pixley-Roy topology: Basic neighborhoods of $F\in F[X]$ are the sets
$$[F,V]=\{H\in F[X]; F\subseteq H\subseteq V\}$$
for open sets $V\supseteq F$, see e.g. here.
 A: The paracompactness of Pixley-Roy Hyperspaces seems to have been solved by Przymusiński (Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, pp.201-219).  Therein he gives the following theorem:
Theorem: The following are equivalent for a Pixley-Roy hyperspace $\mathcal{F}[X]$:


*

*$\mathcal{F}[X]$ is paracompact;

*$\mathcal{F}[X]^n$ is paracompact for all $n \in \mathbb{N}$;

*$\mathcal{F}[X]$ is collectionwise Hausdorff;

*There is a neighbourhood assignment $F \mapsto V_F$ for the finite nonempty subsets of $X$ such that given finite, nonempty $F ,H \subseteq X$ the inclusions $F \subseteq V_H$ and $H \subseteq V_F$ imply $F \cap H \neq \emptyset$.


It follows from this that $\mathcal{F}[\mathbb{R}]$ is not paracompact.  Suppose that $V \mapsto V_F$ is any neighbourhood assignment for the finite nonempty $F \subseteq \mathbb{R}$.  Note that given any $x \in \mathbb{R}$ we may assume without loss of generality that $V_{\{x\}}$ is of the form $( x - \varepsilon_x , x + \varepsilon_x )$ for some $\varepsilon_x > 0$.  From now on, I will write $V_x$ for $V_{\{x\}}$.  The non-paracompactness of $\mathcal{F}[\mathbb{R}]$ is then an immediate consequence of the following claim:
Claim: There are distinct $x, y \in \mathbb{R}$ such that $y \in V_{x}$ and $x \in V_{y}$.
Proof.  We construct sequences $\{ x_i \}_{i \in \mathbb{N}}$ and $\{ \delta_i \}_{i \in \mathbb{N}}$ so that


*

*$x_0 = 0$;

*$\delta_i = \varepsilon_{x_i}$;

*$x_{i+1} = x_i + (-1)^i \frac{\delta_i}{2}$


Note that for $j > i$ we clearly have that $x_j \in V_{x_i}$.  If for any $j > i$ we also have $x_i \in V_{x_j}$ we are done.  So assume that this never happens.
Given any $i \in \mathbb{N}$, we have $x_i \notin V_{x_{i+1}}$.  It thus follows that $$| x_{i+1} - x_i | > \varepsilon_{x_{i+1}} = \delta_{i+1} = 2 | x_{i+2} - x_{i+1} |,$$  and so the sequence $\{ x_i \}_{i \in \mathbb{N}}$ converges to some $y \in \mathbb{R}$.
It can be shown that $y \in V_{x_i}$ for all $i \in \mathbb{N}$.  (If $i$ is even, then for all $j > i+1$ we have $x_i < x_j < x_{i+1}$ and therefore $x_i \leq y \leq x_{i+1}$; similarly for $i$ odd.)  As $x_i \rightarrow y$, then $| y - x_i | < \varepsilon_y$ for all large enough $i$, and therefore $x_i \in V_y$ for all such $i$. $\Box$
