Finite union of closed paracompact subspaces is paracompact. This is an excercise from the Munkres, Section 41, excercise 7)a:
If $X$ is a finite union of closed paracompact subspaces of $X$, then $X$ is paracompact.
Well, I tried to prove it for just two, i.e., $X = K_1 \cup K_2$, with $K_1$ and $K_2$ closed paracompact subspaces. Then, by induction, I'll be able to do it for $m$ finite subspaces.
Now, given an open cover of $X$, $\{U_i\}_{i\in I}$, I want to find a refinament locally finite. As both $K_1$ and $K_2$ are paracompacts, then I have $\{V^{1}_j\}_{j\in J}$ and $\{V^{2}_j\}_{j\in J}$ renfiaments locally finite for $K_1$ and $K_2$ respectively. But now I'm lost, I don't know how to use this, or the fact that $K_i$ are closed, maybe I have to define $\{V^{1}_j \cap K_2 \}_{j\in J}$ or something like that.
If you can give me a hand, I'd appreciate it.
Thanks
 A: Let $\mathcal{U}$ be an open covering of a regular space $X=\bigcup_\mathbb{N}X_n$ which is a union of countably many closed paracompact subspaces $X_n\subseteq X$ whose interiors $(X_n)^\circ$ cover $X$.
For each $n\in\mathbb{N}$ the family $\mathcal{U}_n=\{U\cap X_n\}_{U\in\mathcal{U}}$ is an open covering of the paracompact $X_n$. Thus there is a family $\mathcal{V}_n'$ of open subsets of $X$ which has the property that $\{V\cap X_n\}_{V\in\mathcal{V}'_n}$ is a locally-finite open refinement (in $X_n)$ of $\mathcal{U}_n$. Write $\mathcal{V}_n=\{V\cap(X_n)^\circ\}_{V\in\mathcal{V}'_n}$ to obtain a locally-finite family of open subsets of $X$ which covers the interior $(X_n)^\circ$ and refines $\mathcal{U}_n$.
The collection
$$\mathcal{V}=\bigcup_\mathbb{N}\mathcal{V}_n$$
is now a $\sigma$-locally-finite open refinement of $\mathcal{U}$, so we have a locally finite open refinement of $\mathcal{U}$, i.e. is paracompact
A: Recall that if $X$ is regular then the following conditions on $X$ are equivalent:

*

*Every open covering of $X$ has a refinement that is an open covering of $X$ and locally finite.

*Every open covering of $X$ has a refinement that is a closed covering of $X$ and locally finite.

(This is Lemma 41.3 in Munkres)
This means we have two closed locally finite refinements covering $K_1$ and $K_2$, and since $K_1$ and $K_2$ are closed, the union of these coverings are closed in our space $X$. It should be clear that the union of these closed coverings are also a refinement on your open covering on $X$. The union is also locally finite since for each $x$ in $X$ there are neighborhoods $U_1$ and $U_2$ that intersects only finitely many elements of the closed coverings of $K_1$ and $K_2$ respectively hence the open set $U_1 \cap U_2$ intersects only finitely many elements of the closed covering of $X$. Hence we have found a for every open covering of a regular space $X$ a refinement that is a closed covering of $X$ and locally finite. This is equivelent to paracompactness of $X$ by Lemma 41.3. This should easily extend to any finite union of paracompact subspaces without use of induction, the finite intersections of open neighborhoods of $x$ will remain open.
