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.
Yes, an $F_\sigma$ set of the first category is the union of countably many closed nowhere dense sets.
Since an $F_\sigma$ set of the first category is the union of countably many closed sets of the first category, it will suffice to show that a closed set of the first category is the union of countably many closed nowhere dense sets. Well, a closed set of the first category is the union of countably many nowhere dense sets, and it is also the union of the closures of those nowhere dense sets, and the closure of a nowhere dense set is a closed nowhere dense set, Q.E.D.
Actually, a closed set of the first category (in a Baire space) is nowhere dense; so there is a shorter proof, if you're working with a Baire space.
Best Answer
Yes, it can. Call $C$ Cantor's set, which is closed, uncountable and with empty interior. $\Bbb Q\cap C$ is dense in $C$, because it contains all the points of $C$ that have finite ternary expansion. Therefore, $\Bbb Q\cap C$ is an example of what you've asked.
More generally, $\Bbb R$ is a separable metric space, therefore all its subsets have a countable dense subset (i.e. for all $X\subseteq \Bbb R$ there is some countable $S\subseteq X$ such that $\overline S\supseteq X$). Any such subset $S$ coming from a closed uncountable subset $X$ with empty interior will do.