Two subsets of $\mathbb R$ are said to be non-overlapping, if they have no interior points in common.
A collection of non-degenerate closed and mutually non-overlapping intervals $\mathcal C$ is said to be an exact partition of $\mathbb R,$ if $\displaystyle\bigcup\mathcal C=\mathbb R.$
It's easy to check that every exact partition of $\mathbb R$ is a countable set, thus one may be inclined to say that we can always express an exact partition of $\mathbb R$ as the following form: $$\mathcal C=\left\{[x_i,x_{i+1}]|\ i\in\mathbb Z\right\}.\tag{1}$$
So we have the following plausible claim:
Claim 1: Every exact partition of $\mathbb R$ could be expressed in the form $(1).$
If the claim above is correct, then for any $a<b\in \mathbb R,$ only finite intervals in $\mathcal C$ would "fall into" $(a,b].$ ie., the set $\left\{X\in \mathcal C|\ X\subset (a,b]\right\}$ is finite.
Inference of claim 1: If $\mathcal C$ is an exact partition of $\mathbb R,$ then for any $a<b\in\mathbb R,$ the set $\left\{X\in \mathcal C|\ X\subset (a,b]\right\}$ is finite.
However, here comes the counterexample:$(0,1]=\displaystyle \bigcup_{n=1}^\infty \Big[{1\over n+1},{1\over n}\Big],$ while $\mathcal C:=\left\{[n,n+1]|\ n\in \mathbb Z,\ n\leqslant -1\right\}\cup \left\{\Big[\displaystyle{1\over n+1},{1\over n}\Big]\bigg|\ n\in\mathbb N\right\}\cup \left\{[n,n+1]|\ n\in\mathbb N\right\}$ is an exact partition of $\mathbb R.$ So the claim 1 is actually not correct.
Question: What requirements should be added to the definition of exact partition $\mathcal C,$ so that $(1)$ would be true?