Sure.
Let $f:[0, 1] \to [0, 1]$ be such that $$f(0. b_1 b_2 b_3\ldots) = \liminf_{N\to \infty} \frac{1}{N}\sum_{n=1}^N b_n$$ where $\{b_n\}$ is the usual binary representation of a number in that interval. We have that $f$ is well defined, measurable and integrable and the Birkhoff ergodic theorem implies that $\lambda(f^{-1}(1/2)) = 1$. For all other values $t\in [0, 1]$, $A_t = f^{-1}(t)$ has measure $0$.
Furthermore $A_s\cap A_t =\emptyset$ for $s\neq t$, and for any $N$ and $0\leq n < 2^N$, $A_t \cap [\frac{n}{2^N}, \frac{n+1}{2^N}] \neq \emptyset$ (the initial part of the expansion doesn't really matter, so $A_t$ has elements in every interval), and so is dense in $[0, 1]$ for all $t$.
The collection $\{A_t\}_{t\in[0, 1] \setminus 1/2}$ is what you were looking for.
I just read your comment to the other answer. I'm not sure that the sets I have here are Borel. Let me think about that for a bit.
As mentioned in the comments, we have the $\liminf$ of Borel functions so we have a Borel function and it's preimages of singletons are therefore also Borel. I'm satisfied that this is all correct, but I don't have a specific reference, sorry.
Regarding uncountability, there are at least two arguments that come to mind. The first, which I had in mind as I wrote, relates to a class of measures on $\prod_{n\in \mathbb{N}} \{0, 1\}$, namely those measures that correspond to an infinite product of the measure where $\mu(\{0\}) = 1-t$ and $\mu(\{1\}) = t$. This is a non-atomic (for $t\neq 0\text{ or }1$) standard probability space, and so a set with measure $1$, which $A_t$ corresponds after a natural map, must be uncountable. (I really like this argument because it highlights that having measure $1$ is more about conforming to the expectations of a measure than being large).
More directly, assume that we have $A_t = \{a_1, a_2, a_3, \dots\}.$ For some $t$, and that $a_i = 0. b_{i, 1} b_{i, 2} b_{i, 3}\ldots$ where this is the usual binary representation of $a_i$. Using a standard diagonalisation argument would be problematic as we would lose control of the assymptotic density of $1$s in its representation. However, if we form the number $a$ which is such that $a= 0.b_1 b_2 b_3\ldots$ where $$b_i = \begin{cases} 1-b_{n,n^2} &\text{if } i = n^2 \text{ for some } n\in\mathbb{N} \\
b_{1, i}& \text{otherwise.} \end{cases}$$ then I claim that $f(a) = f(a_1)$ as the assymptotic density of the squares is $0$, so $a\in A_t$ but $a\neq a_n$ for all $n$ as their binary representations differ at the $n^2$ place. So $A_t$'s elements cannot be listed and $A_t$ is uncountable.
On open set $S$ you can say that elements $x,y\in S$ are related if the closed interval (or singleton) determined by them is a subset of $S$.
(So $x\sim y$ if $x=y$ or $x<y$ and $[x,y]\subseteq S$ or $y<x$ and $[y,x]\subseteq S$)
This relation can be shown to be an equivalence relation and the equivalence classes can be shown to be open.
Now the density of $\mathbb Q$ comes in in order to prove that there are at most countable equivalence classes.
Best Answer
There is an uncountable set of countable dense sets. Define an equivalence relation on the reals as $x\sim y \iff x-y \in \Bbb Q.\ \ $ Each equivalence class is countable, so there is an uncountable number of them. They are disjoint and dense.