Proposition 16

Every closed subspace $F$ of a paracompact space $X$ is paracompact.

Certainly $F$ is Hausdorff. On the other hand, if $(V_i)$ is an open covering in the subspace $F$, then each $V_i$ is of the form $V_i = U_i \cap F$, where $U_i$ is open in $X$. Consider the open covering $\mathscr{R}$ of $X$ formed by $CF$ and the $U_i$; since $X$ is paracompact, $\mathscr{R}$ has a locally finite refinement $\mathscr{R}'$, and the intersections with $F$ of the sets belonging to $\mathscr{R}'$ form a locally finite open covering of $F$ which is finer than the given covering $(V_i)$.

On the other hand an open subspace of a compact space need not be paracompact (Exercise 11).