Can we find uncountably many disjoint dense measurable uncountable subsets of $[0,1]$? Can we find uncountably many disjoint dense measurable uncountable subsets of $[0,1]$? Obviously we may as well assume all the subsets have measure $0$. If I didn't specify the subsets were uncountable, we could easily find uncountably many dense disjoint measurable subsets $X_\alpha$ (all countable) by taking a Hamel basis $H$ of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, where $H$ contains a non-zero rational number, and then set $X_\alpha = (\mathbb{Q} + \alpha) \cap [0,1]$ for every $\alpha \in H$.
But what about if we want uncountable measurable dense subsets? I think it should still be that there are uncountably many that can be found that are disjoint, according to my intuition, but I don't know an "easy" way to show it like I can in the case of countable dense subsets. 
 A: 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.
A: The following, which is related to the answer user24142 gave, is taken from this 21 January 2003 sci.math post:
[. . .] here's a collection of $c$-many pairwise disjoint sets with cardinality $c$ in every open interval. In fact, each of these sets will have a positive Hausdorff dimension intersection with every open interval and they are not very complicated from a descriptive set theoretic point of view (they're all $F_{\sigma \delta}$ sets, I believe), and thus are very low in the Borel hierarchy of sets.
For each real number $r$ between $0$ and $1$ (inclusive), let $P_r$ be the set of irrational numbers whose decimal expansions have $r$ for their limiting proportion of $5$'s. That is, if $N(x,5,n)$ is the number of appearances of the digit $5$ to the right of the decimal point in the first $n$ digits of the decimal expansion of the real number $x,$ then $P_r$ is the set of real numbers $x$ such that $\lim_\limits{n \to \infty}{N(x,5,n)}$ exists and is equal to $r.$
Then it is known that each $P_r$ is a Borel set (in fact, $F_{\sigma \delta}$ I think) with positive Hausdorff dimension in every open interval (see these sci.math posts: 16 June 2001 and 19 February 2003), which is quite a bit more than simply saying that each open interval has $c$-many numbers belonging to each of the $P_r$ sets.
On the other hand, these sets are small in a couple of other ways. All of them except for $P_{\frac{1}{10}}$ have measure zero, and all of them, including $P_{\frac{1}{10}},$ are first (Baire) category sets. Incidentally, the union of all the $P_r$ sets doesn't include all the real numbers. If you let $Q$ be the remaining real numbers, then $Q$ is a measure zero set with a first category complement. Hence, $Q$ is co-meager in every open interval, and so it too will have $c$-many elements in every open interval.
A: It's enough to do this for $\mathbb R,$ since $(0,1)$ and $\mathbb R$ are homeomorphic. Let $c$ denote the cardinality of $\mathbb R.$
Claim: There are $c$ many disjoint countable dense subsets of $\mathbb R.$
Proof: Let $K$ be the classical Cantor set. By "a Cantor set" I'll mean any set of the form $a+bK,$ where $a\in \mathbb R$ and $b>0.$
Let $I_1,I_2,\dots$ be the open intervals with rational end points. Then we can inductively choose disjoint Cantor sets $K_n\subset I_n$ for $n=1,2,\dots.$ (This is  pretty straightforward. I'll leave it for now; ask questions if you like.)
For each $n,$ write $K_n = a_n + b_nK,$ and let  $f_n: K \to K_n$ be the map $t\to a_n +b_nt.$ For $t\in K,$ define the sets $D_t = \{f_n(t): n=1,2,\dots \}.$ Then the sets $D_t$ are disjoint. Each $D_t$ is dense in $\mathbb R,$ simply because $D_t$ contains a point in each of the intervals $I_n.$ Because $K$ has cardinality $c,$ the claim follows.
To prove the full result, observe that there is a continuous surjective map $g:K\to [0,1]\times [0,1].$ (For example $g =P\circ C,$ where $P$ is the Peano space filling curve, and $C$ is the Cantor function.) For $t\in K,$ let $K_t = \{t\} \times [0,1].$ Then $\{ g^{-1}(K_t): t \in K\}$ is a disjoint collection of $c$ many sets, each of which is compact and of cardinality $c.$
Finally, consider the sets $D_t\cap (\mathbb R\setminus K) \cup g^{-1}(K_t)$ for $t\in K.$ These sets are disjoint, and each such set is dense in $\mathbb R$ and has cardinality $c.$ Since there are $c$ many such sets, we're done.
A: I found a construction based on this idea.
We need the following lemma:
If $M$ is an uncountable set, then there is an uncountable family of pairwise disjoint uncountable subsets of $M$.
To prove this assertion, observe that $|M \times M| = |M|$ and let $f$ be a bijection between the two sets. The family $\{f(\{x\} \times M) \;|\; x \in M\}$ satisfies the conditions of our lemma.
Now we only apply this lemma to the Cantor set to get an uncountable family of disjoint uncountable subsets. Since the Cantor set is a nullset, every subset is a nullset and therefore Lebesgue-measurable. Now we only need to join every set with one of the sets $X_\alpha$ from the starting post to get an uncountable family of uncountable disjoint dense subsets of $[0, 1]$.
Edit: I think the sets can be chosen Borel with the help of Zorn's Lemma, but I can't quite finish the proof. Let $B$ be the set of all uncountable Borel-nullsets and define 
$$S = \{M \in \mathcal{P}(B) \;|\; \text{the sets in $M$ are pairwise disjoint}\}.$$
$S$ is obviously nonempty and we can define a partial order on $S$ via inclusion. Now every chain in S has an upper bound, namely the union of all elements in the chain. Therefore by Zorn's lemma there is a maximal element $T \in S$. We are done if we can prove that $T$ is uncountable.
Let's assume that $T$ is countable and that $N$ is the union of all sets in $T$. Then $N$ is Borel-measurable and a nullset. We now only need to prove that there is an uncountable Borel-nullset in $[0, 1] \setminus N$, but I can't find an argument to prove this assertion.
