Every closed subspace of a paracompact space $X$ is paracompact.
My attempt:
Let $A\subset X$ be closed and $\{U_{\alpha}\}_{\alpha \in I}$ an open cover of $A$. This means that $U_\alpha = A \cap U_{\alpha}^{'}$ for some open subset $U_{\alpha}^{'} \subset X. $Then the collection $\mathscr{U} = \{X \setminus A\} \cup \{U_{\alpha}^{'}\}_{\alpha \in I}$ is a cover of $X$. Since $X$ is paracompact, there is a locally finite open refinement $\{V_{\beta}\}_{\beta \in J}$ of $\mathscr{U}$ that covers $X$. Let $\mathscr{V} = \{A \cap V_{\beta}\}_{\beta \in J}$.
So, I need to test three things:
(1) $\mathscr{V}$ is nonempty
(2) $\mathscr{V}$ is a locally finite refinement of $\{U_{\alpha}\}_{\alpha \in I}$
(3) $A \subset \bigcup_{\beta \in J}A \cap {V}_{\beta}$
I think an alternative way to write the set $\mathscr{V} $ is as follows:
$\mathscr{V} = \left\{V_\beta \colon V_{\beta} \subset U'_{\alpha} \hspace{0.3cm}\text{for some} \hspace{0.3cm}\alpha \in I \right\}$. Let $W = \bigcup_{V_\beta \in \mathscr{V}}V_{\beta}$, then I must prove that $A \subset W$ and that $\mathscr{V}$ refines $\{U_{\alpha}\}_{\alpha \in I}$.
How do I deduce that from what I already have above?
I need some help to do this.
Here are some definitions:
Definition.
Let $X$ be a topological space. A collection of sets ${U_{\alpha} \subset X}$ (not necessarily open or closed) is said to be locally finite if to each ${x \in X}$, there is a neighborhood ${U}$ of ${x}$ that intersects only finitely many of the ${U_{\alpha}}$
Definition. Let ${\left\{U_{\alpha}\right\} }$ be a cover of a space ${X}$. Then a cover ${\left\{V_{\beta}\right\}}$ is called a refinement if each ${V_{\beta}}$ sits inside some ${U_{\alpha}}$
Definition. A Hausdorff space is paracompact if every open covering has a locally finite refinement.
Best Answer
Every $a \in A$ is in some $V_\beta$ (as these form a cover of $X$). But then $a \in V_\beta \cap A \in \mathcal{V}$. These trivial facts prove two things: $\mathcal{V}$ is non-empty if $A$ is (this is a necessary condition for (1) to hold of course), and $\mathcal{V}$ is a an open cover of $A$. (so 1 and 3 hold).
$\mathcal{V}$ refines $\{U_\alpha\}_{\alpha \in I}$: let $V_\beta \cap A$ be an arbitrary non-empty member of $\mathcal{V}$. Then $V_\beta \subseteq U'_\alpha$ for some $\alpha \in I$ (it cannot sit inside $X\setminus A$ because then its intersection with $A$ is empty). So $V_\beta \cap A \subseteq U'_\alpha \cap A=U_\alpha$, and we're done.
$\mathcal{V}$ is locally finite. Let $a \in A$. Then we can find a neighbourhood $N_a$ (in $X$) of $a$ so that $N_a$ intersects at most finitely many members of $\{V_\beta\}_{\beta \in J}$. But then $N_x \cap A$ also intersects at most finitely many sets (as many or fewer than before) of the form $V_\beta \cap A$ ($\beta \in J$) and this is a neighbourhood of $a$ in $A$. So $\mathcal{V}$ is locally finite.
There's no more to it than that. Besides remarking that a subspace of a Hausdorff space is also Hausdorff, but you rightly focused on the cover aspect of the definition.
Notice the similarity to the proof that a closed set of a compact space is again compact.