Can any subset of the unit interval be decomposed into a countable number of everywhere-dense sets?

Question

Suppose we have an uncountable set $U$ of real numbers in $[0, 1)$.

I would like to decompose it into a family of subsets $S_i$ such that:

• Each of $S_i$ is a dense interval, i.e. is everywhere dense in $[min(S_i), max(S_i))$
• Each of $S_i$ is maximal in this sense, i.e. adding any number from $U \backslash S_i$ would make it stop being everywhere dense in $[min(S_i), max(S_i))$

Is it true that, for any set $U$, there exists such a decomposition into a countable number of subsets $S_i$?

(I've looked at some similar questions, e.g. Topological spaces having countable dense sets but didn't find anything directly addressing my question, and my knowledge of topology is very basic and is not good enough to know what else to search for)

Answer 1

I just realized that this is clearly false: the Cantor set is uncountable and nowhere dense, i.e. it can not be decomposed into everywhere-dense subsets.

(which raises a separate question: is there a name for the property "can be decomposed into a countable number of everywhere dense sets"?)