Every $F_\sigma$-set in a paracompact space is paracompact. 
Every $F_\sigma$-set in a paracompact space is paracompact.

Definitions:

$F_\sigma$-set is a countable union of closed sets
paracompact: if every open cover has an open refinement that is locally finite

Let $(X,\tau)$ be a paracompact topological space and $Y = \cup_{n=1}^{\infty}F_n$ be an $F_\sigma$ subset of $X$. Then for any open cover $\mathcal{A}$ of $Y$, there exists a refinement $\mathcal{W}_n$ of $\mathcal{A}$, that is locally finite on $F_n$. Then the union $\mathcal{W} = \cup_{n=1}^{\infty}\mathcal{W}_n$ will be a refinement of $\mathcal{A}$ and a cover of $Y$, but will it be locally finite? If there is a point $x \in Y$, s.t. $x \in F_n$, for each $n \in \mathbb{N}$?
 A: A classic theorem in General Topology is the following characterisation of paracompactness in regular spaces:
Let $X$ be a regular space. Then TFAE (note: a refinement must itself also be a cover):


*

*Every open cover of $X$ has a locally finite open refinement. (i.e. $X$ is paracompact)

*Every open cover of $X$ has a $\sigma$-locally finite open refinement (where a family of sets is  $\sigma$-locally finite iff it is a countable union of locally finite families)

*Every open cover of $X$ has a locally finite refinement (of any kind).

*Every open cover of $X$ has a locally finite closed refinement.


See Engelking, "General Topology" (2nd ed. thm 5.1.11); Munkres, "Topology" (first ed. chapter 6, lemma 4.4), (2nd edition: lemma 41.3 page 254), Kelley (chapter 5, Thm 28) etc. I also wrote its proof (somewhat sloppily) in my note here.
Then suppose that $X$ is paracompact and Hausdorff, and let $S = \bigcup_n F_n$ (all $F_n$ closed in $X$ of course) be an $F_\sigma$ subset of $X$. Then $S$ is also paracompact (in the subspace topology). 
Proof: let $\mathcal{U} = \{U_i : i \in I\}$ be an open cover of $X$ by relatively open subsets. So we have for each $i$ an open subset $V_i$ of $X$ such that $V_i \cap S = U_i$. Now fix $n$ and consider the open cover $\{V_i: i \in I\} \cup \{X\setminus F_n\}$ of $X$. By paracompactness this has a locally finite refinement $\mathcal{W}_n$. Define $$\mathcal{W'}_n = \{S \cap W: W \in \mathcal{W}_n, W \cap F_n \neq \emptyset\}$$
and note that $\mathcal{W}' = \bigcup_n \mathcal{W'}_n$ is a $\sigma$-locally finite open (in $S$) refinement of $\mathcal{U}$ and as a paracompact Hausdorff space is regular and regularity is hereditary, we conclude by the lemma that $S$ is paracompact.
Note however that by using the fundamental lemma I committed myself to work in regular (and thus even normal) spaces (as paracompact plus Hausdorff implies normal). Also note that we could have started with a regular paracompact (not necessarily $T_1$) space $X$ as well.
This however seems essential: this post by Brian M. Scott gives an example of a compact (non-Hausdorff and non-regular), so paracompact, space with an $F_\sigma$ subspace $Y$ that is not paracompact.
