Covering a union of closed intervals by its subcollection Let $\mathcal{I}=\{I_k\}_{k=1}^N$ be a finite collection of bounded, closed intervals in $\mathbb{R}$. I want to show that there exists a subcollection of $\mathcal{I}$, $\{I_{k_{\ell}}\}\subset \mathcal{I}$, such that
\begin{align*}
 \bigcup_k I_k \subset \bigcup_{\ell} I_{k_{\ell}}
 \end{align*}
and for every $x \in \bigcup_k I_k$, there at most two intervals in $\{I_{k_{\ell}}\}$ to which $x$ belongs. Is this has something to do with the Vitali Covering lemma?
 A: No need to confuse the issue with a fog of subscripts. In plain language the question is, given a finite collection $\mathcal I$ of (bounded) closed intervals of $\mathbb R$, is there a subcollection $\mathcal J\subseteq\mathcal I$ such that $\bigcup\mathcal J=\bigcup\mathcal I$ and no point belongs to more than two elements of $\mathcal J$? The answer is yes. Let $\mathcal J$ be a minimal subcollection of $\mathcal I$ with the same union. I claim that no point is covered by more than two elements of $\mathcal J$.
Consider any point $x\in\bigcup\mathcal J$. The set $F=\bigcup\{J\in\mathcal J:x\in J\}$ is a closed interval containing $x$; say $F=[a,b]$, $a\le x\le b$. Then $[a,x]\subseteq J_1$ for some interval $J_1\in\mathcal J$, and $[x,b]\subseteq J_2$ for some interval $J_2\in \mathcal J$.
Assume for a contradiction that there is an interval $J\in\mathcal J\setminus\{J_1,J_2\}$ with $x\in J$. Then $J\subseteq[a,b]\subseteq J_1\cup J_2$, so $\bigcup(\mathcal J\setminus\{J\})=\bigcup\mathcal J=\bigcup\mathcal I$, contradicting the assumed minimality of $\mathcal J$.
