Can any subset of $\mathbb{R}$ be generated from open intervals?

Not every subset of $\mathbb{R}$ can be "created" by this process. If it were possible, we would get that the Borel $\sigma$-algebra would contain all subsets of $\mathbb{R}$, which it does not.


______Editing my comment to extend the answer_______________

Hint: There is an example for Lebesgue but not Borel measurable sets.

Additionally:-

If you're familiar with measure theory, then there's a couple results that are somewhat related to this matter.

  1. Every non-empty open subset $U \subseteq \mathbb{R}$ is a disjoint union of open intervals.

  2. Littlewood's first principle of Analysis

Every measurable set $E\subset \mathbb{R}$ with $m(E)<$ $\infty$ is almost a finite union of intervals.