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?
Best Answer
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.