Is there an explicit example of an uncountable collection of pairwise disjoint, dense, countable subsets whose union equals $[0,1]$? Question:

Is there an explicit example of an uncountable collection of pairwise
disjoint, dense (in $[0,1]$), countable subsets (i.e. each subset has countable many elements) of $[0,1]$ whose union equals $[0,1]$?

I know that if we change the question to, "Is there an explicit example of an uncountable collection of pairwise disjoint, dense, uncountable subsets of $[0,1]$ ? ", then there is a good answer here.
But I can't seem to answer the question if countable subsets are required. The obvious attempts are something like
$$\huge{\cup}_{\normalsize{\text{p prime}}} \ \normalsize{ \left(\sqrt{p}+\mathbb{Q}\right)}$$
but I think this is the wrong route to go down, because we're not going to get an uncountable number of disjoint sets whose union is $\ [0,1]$.
Alternatively if we start with,
$$\huge{\cup}_{\normalsize{ i\text{ is Irrational}}} \ \normalsize{ \left(i+\mathbb{Q}\right)}$$
then I don't see an easy way to deal with repeat sets, other than invoking the axiom of choice. But I don't want to invoke AOC, I want an explicit example. And repeat sets are a problem because I require them to be disjoint. So I don't see a way round this, although maybe there is one? But probably a different approach altogether is necessary.
Something tells me the answer is easy and I have forgotten the solution, but I can't think of what it is. Also, I require an explicit example: invoking the axiom of choice is not an acceptable answer imo. And I'm fairly sure the answer is "yes", although maybe the answer is no because of something like the Baire Category Theorem, which I don't really understand?
It's also possible this question is not easy (for me) and is out of my depth... I don't know.
 A: Your second example actually doesn't involve AC at all! The set $S$ of shifts of $\mathbb{Q}$ is a perfectly nice object. Picking elements of the elements of that set uses choice, but $S$ itself isn't problematic.
More generally, any countable dense equivalence relation (that is, equivalence relation all of whose classes are countable and dense) on $[0,1]$ will give rise to an example, namely the set of all its classes. Above we're using the "rational distance" equivalence relation, but there are plenty of others (e.g. Turing equivalence).
Choice comes back into the picture if we ask for our collection of sets to be indexed by $\mathbb{R}$. This doesn't quite amount to picking an element of each set - there's no rule saying that the index of $A$ has to be an element of $A$ - but it still makes things much harder. I believe that the theory $\mathsf{ZF+DC+AD}$ (a very well-behaved alternative to $\mathsf{ZFC}$) in fact rules out any possible example - that is, that $\mathsf{ZF+DC+AD}$ proves that there is no family $(A_r)_{r\in\mathbb{R}}$ of subsets of $[0,1]$ such that each $A_r$ is countable and dense in $[0,1]$, $A_r\cap A_s=\emptyset$ whenever $r\not=s$, and $\bigcup_{r\in\mathbb{R}}A_r=[0,1]$ - but I don't immediately see a proof of this.
A: Your issue arises because you want to index every set in the family by one of its elements (ie using a choice function), while this is not technically required by the current formulation of your question. For example, consider :
$$\mathcal F = \Big\{ A \subset [0,1] \Big|\forall x\in A, \forall y \in [0,1], x-y \in \mathbb Q \Leftrightarrow y\in A \Big\}$$
Then $\mathcal F$ is an uncountable partition of $[0,1]$ into disjoint countable subsets. It might not be possible to find a choice function for $\mathcal F$ without AC, although I am really not well versed enough in logic to prove this.
