Partitioning $\mathbb{R}^2$ into disjoint path-connected dense subsets

First we partition $\mathbb Q^2$ into $n$ dense subsets $A_1, \ldots , A_n$ and fix an enumeration $a_{i,1}, a_{i,2}, a_{i,3}, \ldots $ for each of set $A_i$.

As a starter, assign colour $i$ to $a_{i,1}$, that is let $S_i^{(1)}=\{a_{i,1}\}$.

Assume we have the following situation:

$m\in\mathbb N$ and we have pairwise disjoints sets $S_1^{(m)}, \ldots , S_n^{(m)}$, where each $S_i^{(m)}$ is a polygonal curve from $a_{i,1}$ to $a_{i,m}$ with $S_i^{(m)}\cap \mathbb Q^2 = \{a_{i,k}\mid k\le m\}$.

Then the complement $\mathbb R^2\setminus \bigcup S_i^{(m)}$ is connected and open. For $i=1, \ldots , n$ we will append a path to $a_{i,m+1}$ as follows: First observe that the set $$X=\mathbb R^2\setminus\left(\bigcup_{j<i}S_j^{(m+1)}\cup \bigcup_{j\ge i}S_j^{(m)}\right)$$ is open and connected. There is a path in $X\cup\{a_{i,m}\}$ from $a_{i,m}$ to $a_{i,m+1}$. Since $X$ is open, we can replace the path with a polygonal path. Also, we can slightly move the nodes of the path (apart from start and end node) such that all line segments avoid rational points (apart from $a_{i,m}$ and $a_{i,m+1}$); in short, this is possible because we have uncountably many choides to position the nodes in an open disc, but there are only countably many points to avoid. Then let $S_i^{(m+1)}$ be the polyline $S_i^{(m)}$ together with the polyline just found. Especially, $S_i^{(m)}\subset S_i^{(m+1)}$.

When we have done this for all $i$, we have obtained the situation described above, but with $m$ replaced by $m+1$.

Now let $$S_i=\bigcup_{m\in\mathbb N}S_i^{(m)}.$$ Then $S_i$ is pathwise connected (any two points are connected in some $S_i^{(m)}$) and is dense in $\mathbb R^2$ because $A_i\subset S_i$. However, I must admit that we do not have $\mathbb R^2=\bigcup S_i$ yet, i.e. the set $$Y=\mathbb R^2\setminus\bigcup S_i$$ need not be empty.