It seems an path-connected anti-convex subset of $\mathbb{R}^2$ containing $(\mathbb{R}\setminus\mathbb{Q})^2$ exists.

Firstly, let $A$ be a countable, dense subset of $\mathbb{R}^2$, and let $B$ be the union of a family of arcs $\gamma_{a,a'}$, for $a\neq a'\in A$, such that $\gamma_{a,a'}$ is an arc from $a$ to $a'$ with diameter $\leq 2d(a,a')$.

Note that any subset of $\mathbb{R}^2$ containing $B$ is path-connected (you can use the arcs in $B$ to go from any point of $\mathbb{R}^2$ to any other point). Also note that the set $(\mathbb{R}\setminus\mathbb{Q})^2$ is obtained by removing from $\mathbb{R}^2$ a countable sequence of lines, which I will call $(l_n)_{n\in\mathbb{N}}$, such that every nontrivial segment in $\mathbb{R}^2$ intersects infinitely many of the lines $l_n$.

In the following, for any points $x,y\in\mathbb{R}^2$, I will say $\gamma$ is an `arc of circle' from $x$ to $y$ if $\gamma$ is the shorter of the two arcs going from $x$ to $y$ inside some circle containing $x,y$ (in particular, $\gamma$ has diameter $\leq2d(x,y)$).

**Claim:** If $A$ is a dense set of $\mathbb{R}^2$, there is a family of arcs of circle $\gamma_{a,a'}$ from $a$ to $a'$, with $a\neq a'\in A$, such that if $B=\bigcup_{a,a'}\gamma_{a,a'}$, then $B\cap \cup_nl_n$ has no three collinear points, except if all three inside one of the lines $l_n$.

If we prove the claim we will be done: indeed, the set $S:=B\cup(\mathbb{R}\setminus\mathbb{Q})^2$ is path connected, as we noted above, and no segment $[x,y]$ can be contained in $S$; this is obvious if $[x,y]$ is contained in one of the lines $l_n$ (then its intersection with $S$ is countable), and if not, then $[x,y]$ contains infinitely many points outside $(\mathbb{R}\setminus\mathbb{Q})^2$, and $B$ contains at most $2$ points of $[x,y]$.

To prove the claim, let $((a_n,a_n'))_{n\in\mathbb{N}}$ be a numbering of the pairs of points $a,a'$ in $A$ with $a\neq a'$, and we construct the arc $\gamma_{a_n,a_n'}$ recursively. 

In the step $n$, we want to choose an arc of circle $\gamma_n$ from $a_n$ to $a_n'$ such that the points of $\gamma_n\cap\cup_nl_n$ are not collinear (collinear inside some line which is not one of the $l_n$) with points in $C:=\left(\cup_{i=1}^{n-1}\gamma_{a_i,a_i'}\right)\cap\cup_nl_n$ (note that $C$ is countable). So let $E$ be the set of points in $\cup_nl_n$ which are collinear (inside some line which is not one of the $l_n$) to two points of $C$. Note that $E$ is countable, as there are only countably many pairs of points in $C$. Note that all arcs $\gamma$ from $a_n$ to $a_n'$ except countably many contain any points of $E$; if we choose any of those arcs $\gamma$, then no point of $\gamma\cap\cup_nl_n$ can be collinear with two points of $C$. 

We also want no points of $C$ to be collinear with two points of $\gamma_n\cap\cup_nl_n$; it seems that will also happen for all choices of $\gamma_n$ except countably many. To see why, we only have to prove that for each fixed $m,n\in\mathbb{N}$ and each $c\in C$, the points $c,\gamma_n\cap l_n$ and $\gamma_n\cap l_m$ can only be collinear for countably many choices of $\gamma$. It seems that will be the case except if the point $c$ is collinear with $a_n,a_n'$. But we can avoid that: before starting the induction, choose the set $A$ in such a way that, for all $n$, no two points of $A$ lie inside $l_n$, so that the set $F=(\cup_{a\neq a'\in A}\text{line}(a,a'))\cap\cup_nl_n$ is countable , and then choose the curves $\gamma_n$ so that they contains no points in $F$; thus, no point in $C$ can be aligned with two points in $A$.