Let $A$ a disjoint collection of closed sets in $R$. If $A$ contains a bounded set then there is a disjoint subcollection of $A$ This is problem 2A.5 of Sheldon Axler's Measure, Integration, and Real Analysis book (Dec. 2022 electronic update).
Suppose $\mathcal{A}$ is a set of closed subsets of $\mathbb{R}$ such that $\bigcap_{F\in\mathcal{A}}F=\varnothing$. Prove that if $\mathcal{A}$ contains at least one bounded set, then there exists $n\in\mathbb{Z}^+$ and $F_1,\dots,F_n\in\mathcal{A}$ such that $F_1\cap\cdots\cap F_n=\varnothing$.
Here's my proof attempt: We can choose a set $F_0\in\mathcal{A}$ such that $F_0$ is closed bounded. Let $x\in F_0$, so since $\bigcap_{F\in\mathcal{A}}F=\varnothing$ there exists a set $F_{x}\in\mathcal{A}$ such that $x\notin F_{x}$. Thus $x$ is in $\mathbb{R}\setminus F_{x}$, and note that this set is open. Hence the class $\left\{\mathbb{R}\setminus F_{x}:x\in F_0\right\}$, where each $F_x$ is defined for each $x\in F_0$ as previously showed, is an open cover of $F_0$.
By the Heine-Borel Theorem we can choose a finite subset $\left\{\mathbb{R}\setminus F_{x_i}:x_i\in F_0 \text{ for each } i=1,\dots,n\right\}$ of $\left\{\mathbb{R}\setminus F_{x}:x\in F_0\right\}$ that also covers $F_0$. Then we have $F_0\subseteq\bigcup_{i=1}^{n}\left(\mathbb{R}\setminus F_{x_i}\right)$, which implies:
$$F_0\subseteq\mathbb{R}\setminus\left(\bigcap_{i=1}^{n}F_{x_i}\right).$$
Hence $F_0\cap F_{x_1}\cap\dots\cap F_{x_n}=\varnothing$ as desired.   $\square$
Is this proof right?
Thank you.
 A: I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 5 on p.23 in Exercises 2A in this book.
I think your proof is of course right and it is very elegant.
My proof is the following:

$\mathcal{A}$ contains at least one closed and bounded set $F_1$.
Let $M$ be a positive real number such that $F_1\subset [-M,M]$.
Then, $\mathbb{R}\setminus F_1\supset\mathbb{R}\setminus [-M,M]$.
Since $\cap_{F\in\mathcal{A}} F=\emptyset$, $\cup_{F\in\mathcal{A}} \mathbb{R}\setminus F=\mathbb{R}$.
So, $\{\mathbb{R}\setminus F:F\in\mathcal{A}\}$ is an open cover of $\mathbb{R}$.
So, $\{\mathbb{R}\setminus F:F\in\mathcal{A}\}$ is also an open cover of $[-M,M]$.
So, by 2.12 Heine-Borel Theorem on p.19 in the book, the open cover $\{\mathbb{R}\setminus F:F\in\mathcal{A}\}$ of $[-M,M]$ has a finite subcover $\{\mathbb{R}\setminus F_2,\dots,\mathbb{R}\setminus F_n,\}$.
Therefore, $(\mathbb{R}\setminus F_1)\cup(\mathbb{R}\setminus F_2)\cup\dots\cup(\mathbb{R}\setminus F_n)=\mathbb{R}$.
So, $F_1\cap F_2\cap\dots\cap F_n=\emptyset$.

