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.
Best Answer
Maybe this is close to your thinking:
Take a set $A\subset \mathbb R$ that is uncountable and take any subset $B\subset A$ that is countable. Then the set $A\setminus B$ is uncountable. Do all examples of uncountable sets look much like this? Is this the only method of finding such examples?
So for example, we know an interval $[a,b]$ is uncountable. So a simple example of a relatively interesting uncountable set is $[a,b]\setminus S$ where $S$ is some set that I can reasonably show is countable (e.g., all rationals, all algebraic numbers, the range of some sequence, etc.)
But there are much more interesting examples of uncountable sets. Every perfect subset of $\mathbb R$ (i.e., every nonempty closed set without isolated points) is uncountable and in fact has cardinality $\mathfrak c$ of the continuum. All of the Cantor-like sets (including the orginal Cantor ternary set) are perfect and so uncountable.
For this see: Proof that a perfect set is uncountable
If you have studied Lebesgue measure on the real line then you have an abundance of interesting uncountable sets. If $E\subset \mathbb R$ is a measurable set of positive Lebesgue measure then it is trivial to show that $E$ is uncountable. This cheats a bit since every such set $E$ contains a perfect set so we are back to saying it is both uncountable and has cardinality $\mathfrak c$.
Of course there are plenty of other ways an uncountable set might arise naturally. For example let $f:[a,b]\to \mathbb R$ be a continuous function with the property that every level $$f^{-1}(L)= \{x\in [a,b]: f(x)=L\}$$ is countable (i.e., finite or countably infinite). Then there is an uncountable set of points where the derivative $f'(x)$ must exist.
It follows, too, that if $f$ has a derivative at only a countable number of points then there are uncountably many values of $L$ so that the levels $f^{-1}(L)$ are themselves uncountable. (By the way, this leads us back to perfect sets: levels of a continuous function are always closed and most continuous nowhere differentiable functions must have most levels perfect.)
In short you will, before long, see more instances of uncountable sets than you might care to confront.