Prove that if $I$ is an interval of $\mathbb R$, covered by an infinite family $\{I_n, n\in \mathbb N\}$ of open intervals, then (where $\ell(I)$ denotes the length of $I$):
$$\ell(I) \le \sum_{n=1}^{\infty} \ell(I_n)$$
It is very intuitive but hard to prove (in a formal manner). The exercise gives the hint: prove that $\ell(I) = \sup\{\ell(K), \text{$K$ is a compact subset of $I$}\}$. I proved this, but I can't see how this hint makes the problem any easier to prove.
I thought about doing the following: If $I$ is unbounded, then we are done, so let's assume that it's bounded. Due to the connectedness of $I$, we can WLOG suppose that the $I_n$'s are such that $I_n \cap I_{n+1} \neq \varnothing$ for each $n$ (this would be the ideal situation, I guess). By the axiom of countable choice, we can find a sequence $(a_n)$ where $[a_n, a_{n+1}] \subset I_n$ for each $n$. Then:
$$\sum_{n=1}^{\infty} \ell(I_n) = \sum_{n=1}^{\infty} \sup \{ \ell([a_n,a_{n+1}]) = a_{n+1} - a_n\}$$
Does this make any sense? How to proceed? Thank you.
Following the hint let $K\subseteq I$ be compact. As the $I_n$ form an open cover of $K$, there exists a finite subcover. Assume two of these finitely many intervals intersect, say $I_i=(a_i,b_i)$ intersects $I_j=(a_j,b_j)$. Then $I_i\cup I_j$ is itself an open interval $(\min\{a_i,a_j\},\max\{b_i,b_j\})$ of length $$ \max\{b_i,b_j\}-\min\{a_i,a_j\}< b_i+b_j-a_i-a_j=\ell(I_i)+\ell(I_j)$$ Thus if we replace $I_i,I_j$ with this one larger interval, we decrease the number of intervals needed to cover $K$ while at the same time decreasing the sum of their lengths. We repeat this process until we cannot continue, i.e., until ther are no overlapping open intervals. As $K$ is connected, this means that we are left with a single open interval $I^*$. Then $$\ell (K)<\ell(I^*)\le \sum_{i=1}^N \ell(I_i)\le \sum_{i=1}^\infty \ell(I_i).$$ As we can make $\ell(I)-\ell(K)$ arbitrarily small, we conclude that $$\ell (I)\le \sum_{i=1}^\infty \ell(I_i).$$