Separability: Does it require that the finite union of e.g. open intervals are non-disjoint?
E.g. when proving that $L_1([0,1])$ is separable.
Intuitively the open intervals themselves are dense, since for each pair of rationals, there's always more between them.
But what about finite unions of them? Intuitively if the unions were non-disjoint, then the resulting set would be dense. However, when constructing dense sets, I've seen the opposite, that they may not explicitly require non-disjointness.
But how are finite disjoint unions of open intervals with rational endpoints dense?