Every closed subspace of a paracompact space is paracompact

For each member of $\mathcal{U}$, take all ambient open $O$ whose trace $O\cap A$ is that member; together with $X\setminus A$, these ambient open sets form an open cover $\mathcal{W}$ of $X$.

By [F2], fix a locally finite open cover $\mathcal{V}$ refining $\mathcal{W}$.

The nonempty traces $V\cap A$ for $V\in \mathcal{V}$ cover $A$, are open in $A$, and refine $\mathcal{U}$: a $V$ meeting $A$ cannot be contained in $X\setminus A$, so its containing member of $\mathcal{W}$ is an ambient representative of a member of $\mathcal{U}$.

These traces are locally finite in $A$, because the trace on $A$ of a neighbourhood in $X$ meeting only finitely many $V$ meets only the corresponding finitely many traces.

The family in step 2.1 is therefore the locally finite open refinement required for $A$, so $A$ is paracompact.