TITLE: A partition of the unit interval into uncountably many dense uncountable subsets
QUESTION [11 upvotes]: The title says it all: Is there a partition of $[0,1]$ into uncountably many dense uncountable subsets ?

REPLY [4 votes]: Here is another solution: An $x\in[0,1]$ is irrational iff its continued fraction is infinite. Recall that the continued fraction of $x$,
 $$ x=\frac1{a_0+\frac1{a_1+\frac1{\dots}}}, $$
is obtained by letting $a_0$ be the integer part of $1/x$, then letting $a_1$ be the integer part of $\displaystyle \frac1x-a_0$, etc, so each $a_i$ is a positive integer.
Recall now that the set $(\mathbb Z^+)^{\mathbb N}$ of infinite sequences of positive integers is uncountable (in fact, $|(\mathbb Z^+)^{\mathbb N}|=|\mathbb R|$). Each such sequence has the form $(b_0,b_1,b_2,\dots)$ with each $b_i$ a positive integer.
Now, for your partition: Given $t\in(\mathbb Z^+)^{\mathbb N}$, put $x\in[0,1]$ in $A_t$ iff, with the $a_i$ as above, we have that $(a_0,a_2,a_4,\dots)=t$. Note that each $A_t$ is closed, uncountable, and they are pairwise disjoint. The union of all the $A_t$ is $[0,1]\setminus\mathbb Q$. Pick countably many $t$, and add an element of $[0,1]\cap\mathbb Q$ to each of these $A_t$. The resulting sets are again closed.
To further satisfy the requirement that the sets are dense, let me modify the construction somewhat: Say that two such sequences $(a_0,a_1,\dots)$ and $(b_0,b_1,\dots)$ are $E$-equivalent iff there are $n,m$ such that $b_{n+k}=a_{m+k}$ for $k=0,1,\dots$, that is, iff, by chopping off initial segments from each (possibly of different length) we obtain the same sequence. This is an equivalence relation on $(\mathbb Z^+)^{\mathbb N}$. 
Noting that each equivalence class is countable, it follows from the fact that $|(\mathbb Z^+)^{\mathbb N}|=|\mathbb R|$ that there are as many equivalence classes as there are reals. Pick from each equivalence class a representative, and call $V$ the set of representatives so chosen, so $V\subseteq (\mathbb Z^+)^{\mathbb N}$ meets each equivalence class in exactly one point.
Now, for your partition:  Given $t\in V$, put $x\in[0,1]$ in $B_t$ iff, with $$ x=\frac1{a_0+\frac1{a_1+\frac1{\dots}}} $$ as above, we have that $(a_0,a_2,a_4,\dots)\mathrel{E} t$. Note that each $B_t$ is $F_\sigma$, uncountable, and dense, and again they are pairwise disjoint. The union of all the $B_t$ is again $[0,1]\setminus\mathbb Q$. Since $[0,1]\cap\mathbb Q$ is itself an $F_\sigma$ set, we can pick a $t_0\in V$ and add to $B_{t_0}$ the set $[0,1]\cap\mathbb Q$, and the resulting set is once again $F_\sigma$. This is of course optimal in the sense that we cannot obtain such a partition using closed sets, because of the density requirement.