Partitions of $\mathbb{R}^2$ into disjoint, connected, dense subsets. Does there exist pairwise disjoint, connected, dense subsets $U_1,\dots, U_n \subset \mathbb{R}^2$ such that $U_1\cup \cdots \cup U_n =\mathbb{R}^2$? 
If $n=1$, then we can take $U_1 = \mathbb{R}^2$. 
If $n=2$, then we can take
$$U_1= \{(x,y)\in\mathbb{R}^2| y\neq 0~\text{and} \sqrt{x^2+y^2}\in\mathbb{Q}\}\cup\{(x,0)\in\mathbb{R}^2|x\geq 0\}$$
and
$$U_2=\{(x,y)\in\mathbb{R}^2|y\neq 0~\text{and}~\sqrt{x^2+y^2}\notin\mathbb{Q}\}\cup\{(x,0)\in\mathbb{R}^2|x<0\}.$$
I do not know how to construct such sets $U_1,\dots, U_n$ for $n\geq 3$, nor do I know a proof that it is impossible. 
Edit: Lukas showed below how to construct such sets $U_1,\dots, U_n$. What if instead of insisting that each $U_i$ is connected, we insist that each $U_i$ is path connected?
 A: Here is a construction in the $2$-sphere $S^2$, equipped with any reasonable metric. By removing one point it becomes homeomorphic to the plane, so it gives an example in $\mathbb{R}^2$. (You have to be a little careful which point to remove, but it is not that hard to figure out that there exists one that works. Alternatively, equip $\mathbb{R}^2$ with a bounded metric and run the same construction.) The construction is similar to the standard "Lakes of Wada" construction in spirit.
Let $U_1^1$ be a simple path which is $1$-dense in $S^2$, i.e., such that every point on the sphere has distance $\le 1$ to a point on $U_1^1$. Now let $U_2^1$ be a simple path (i.e., a homeomorphic image of $[0,1]$) in $S^2 \setminus U_1^1$ which is $1$-dense in $S^2$. Proceed to get disjoint $1$-dense simple paths $U_1^1,\ldots U_n^1$. Now extend $U_1^1$ to obtain a $1/2$-dense simple path $U_1^2$ in $S^2 \setminus (\bigcup_k U_k^1)$. Inductively construct a sequence of mutually disjoint simple paths $U_1^2,\ldots U_n^2$ which are $1/2$-dense extensions of $U_1^1,\ldots,U_n^1$. Now keep extending those inductively to get mutually disjoint paths $U_1^m,\ldots U_n^m$ which are $1/m$-dense in $S^2$. This construction is possible because at any step the complement of the already constructed paths is connected, since it is the complement in $S^2$ of a finite set of disjoint homeomorphic images of $[0,1]$.
Now let $U_k^\infty = \bigcup_m U_k^m$ for $k=1,\ldots,n$. This is a collection of mutually disjoint open paths (continuous images of $[0,1)$ or $(0,1)$, depending on how exactly the extensions are chosen), each of them dense in the plane. Their union is not necessarily all of $\mathbb{R}^2$, so let $T=S^2 \setminus \bigcup_k U_k^\infty$, and let $U_1 = U_1^\infty \cup T$ and $U_k = U_k^\infty$ for $k\ge 2$. Then $S^2 = \bigcup_k U_k$ is a disjoint partition, and since $U_2,\ldots,U_n$ are continuous images of an interval, they are connected, even path-connected. The set $U_1$ is not necessarily path-connected, so in order to show connectedness assume that $U_1 = A \cup B$ with relatively open disjoint sets $A$ and $B$. Since $U_1^\infty$ is path-connected, it has to be contained in either $A$ or $B$. We may assume $U_1^\infty \subseteq A$. Assume $t \in T \cap B$. Since $U_1^\infty$ is dense and $B$ is relatively open, there has to exist $u \in U_1^\infty \cap B$. However, this contradicts $A \cap B = \emptyset$.
The last argument is probably some standard topology result, that if $U$ is connected, and $V\supseteq U$ is contained in the closure of $U$, then $V$ is connected. The crucial point is to find disjoint connected dense subsets in the first place.
This construction does not guarantee that $U_1$ is path-connected, and I am not sure whether the similar question about a path-connected partition has a positive answer.
A: Here's an explicit construction that I think will work.  This is for the connected, not the path connected, case.
Take $X_1$, ..., $X_n$ to be a partition of $\mathbb{R}$ into dense subsets.  (For example, for $i > 1$ choose a prime $p_i$ and take $X_i$ be all rational numbers with denominator of the form ${p_i}^j$, and finally take $X_1$ to be the complement.)  View this as a function $f: \mathbb{R} \rightarrow \{1, ..., n\}$.  Take $Y_1$, ..., $Y_n$ to be a partition of $\mathbb{R}$ into intervals.  Then define $U_i = \bigcup_{x \in \mathbb{R}} \{x\} \times Y_{f(x)}$.
It's clear that the $U_i$ form a partition and that each one is dense.  I am not sure how to show that each one is connected, but here's a sketch.  A disconnection of $U_i$ would have the form $A_1$, $A_2$, where the $A_i$ are disjoint open sets which cover $U_i$.  In this case I would expect there is an embedded circle $C \subseteq \mathbb{R}^2 - A_1 \cup A_2$.  (The circle might pass through the point at infinity.)  In particular $C$ lies in the complement of $U_i$.  But the complement of $U_i$ has the same form as $U_i$ itself.  Its path components are straight lines, and there is no room there for the circle $C$.
Here are crude pictures with each $U_i$ in a different color.  The original path-connected construction with $n = 2$ looks like 
The new construction with $n = 3$ looks like 
A: UPDATED 2012-12-15
For the path-connected case, the original post shows that we can achieve $1$ and $2$.  I claim that we can also achieve continuum-many.  I'd still like to know about all the cardinalities in between.
Claim.  There is a solution with continuum-many path-connected $U_i$.
Proof.  The basic idea is this: since the original construction is a one-dimensional foliation almost everywhere, what if we try to make it a foliation?
Start with $S^2$ rather than $\mathbb{R}^2$.  Take the northern hemisphere $H_0$, whose boundary is the equator $S^1$.  Suppose $f_0: S^1 \rightarrow S^1$ is a reflection.  Pick a point $x \in S^1$.  Define $p_0$ as the line segment from $x$ to $f_0(x)$, projected straight up to $H_0$.  $p_0$ is a path properly embedded in $H_0$, although it's degenerate when $f_0(x) = x$.  As $x$ varies, the different paths $p_0$ give a partition (I think it's called a foliation) of $H_0$.
Suppose $f_1$ is also a reflection of $S^1$.  Use that similarly to define a path $p_1$ in the southern hemisphere $H_1$ starting at $p_0(x)$.  Define $f = f_1 \circ f_0$, which is a rotation of $S^1$, and $p$ as the path composition $p_0 p_1$.  Then $p$ is a path in $S^2$ from $x$ to $f(x)$ which meets $S^1$ at $x$, $f_0(x)$, and $f(x)$.  Repeat this step to extend $p$ from $[0, 1]$ to $[0, \infty)$ so that $p(i) = f^i(x)$ for all integers $i \ge 0$.  Since $f$ is a homeomorphism of $S^1$, we can also run the process backwards to extend $p$ to a path $\mathbb{R} \rightarrow S^2$ with $p(i) = f^i(x)$ for all integers $i$.
Now let $x$ vary over $S^1$, giving us paths $p_x$.  If $A_x$ is the orbit of $x$ under the action by iterations of $f$, then the intersection of $p_x$ and $S^1$ is $A_x \cup f_0(A_x)$.  For any point $y$ in that intersection, $p_y$ is basically the same as $p_x$: the domain has been translated, but the image is the same.  So if we define $U_x = p_x(\mathbb{R})$, the choice of $f_0$ and $f_1$ gives us a partition $\{U_x\}$ of $S^2$ into paths.
Then of course we choose the $f_i$ so that $f$ is a rotation by a nasty angle $\alpha$ with $\alpha / 2 \pi$ irrational.  Each $A_x$ is dense in $S^1$, and each $p_x$ is dense in $S^2$.  The family $\{U_x\}$ is a partition of $S^2$ into dense path-connected subsets.
The original question was about $\mathbb{R}^2$, so puncture $S^2$ by removing a point.  That cuts one of the $U_x$ in half, but each half is still path-connected and (it's pretty clear) dense.  It's also pretty clear that the cardinality of $\{U_x\}$ is the continuum.  For example, each $U_x$ has measure $0$. QED.
A: Here, I write the answer for the path-connected case. As you mentioned such a partition for 
$n=1, 2$ is possible. But, for any countable cardinal $n\geq 3$ this is not possible. First, remember that a subset of a metric space is said to be of the first category if it is the union of at most countably many nowhere dense sets. Indeed, Mary-Elen Rudin in https://projecteuclid.org/euclid.dmj/1077374790 showed that if there are three disjoint path-connected sets which are
dense in $\mathbb{R}^2$, then one of them is of the first category. Now, as $\mathbb{R}^2$ is not of the first category and a countable union of the first category sets is of the first category, the claim is proved. 
Here https://msp.org/involve/2014/7-2/p01.xhtml, you can see an interesting application of this result which gives a new proof of the non-existence of bijection continuous map between $\mathbb{R}^2$ and $\mathbb{R}^n$ for $n\neq 2$.
A: You can always take a dense partition of $\mathbb R$ in $n$ subsets and consider the radii. For instance, if $V_1, \dots, V_n$ is a dense partition of $\mathbb R$, then 
$$
U_i = \{ (x,y) \in \mathbb R^2 \, | \, \sqrt{x^2 + y^2} \in V_i \} 
$$
is a dense subset of $\mathbb R^2$. It's not connected, but in the same way you used the real line to make $U_1$ and $U_2$ in your case, you can play around with the lines $y = ix$ for instance, to make sure $U_i$ is connected. It would probably look like this : 
$$
U_i = \left( \{ (x,y) \in \mathbb R^2 \, | \, \sqrt{x^2 + y^2} \in V_i, (x > 0 \quad \Rightarrow \quad y \neq jx, j = 1, 2, \dots, i-1, i+1, \dots n,  ) \} \cup \{ (x,y) \in \mathbb R^2 \, | \, y = i x , x > 0\}\right)
$$
and you can put $0 \in U_1$ to make all your conditions satisfied. It's pretty complicated to write down but it's all pretty simple ; I used the partition of $\mathbb R$ to get my partitions of $\mathbb R^2$ dense, and then "connected them" using the lines going to the origin. 
Hope that helps,
