A have a (non-simply) bounded path-connected open set $S\subset\mathbb{R}^2$. 


Given $x\in S$, there are paths in $S$ from $x$ to any point in the boundary $\partial S$. 
However, can all these paths be constructed so that they don't cross?. 


**Note**: If $S$ is simply connected, then by the Riemann mapping theorem, we can map $S$ to the unit disk, and connect $x$ to points in the boundary by (shifted) wheel spokes. 
In the non-simply connected case, the paths have to somehow "avoid" the holes.. 
Perhaps there's some differential equation with whose solution produces such paths?
(maybe something like a laplace equation solution with boundary value f=1 at x and at the holes, and f=0 at the boundary, and paths following the gradient of f..?)



Given $x$ in a path-connected open set $S$ on the plane, are there non-crossing paths from $x$ to every point in $\partial S$?
Consider a double-sided topologist's comb, with teeth $\{1/n\} \times I$ for $n \in \mathbb{Z}, n \ne 0$ together with $[-1,1] \times {0}$ and ${0} \times I$. This is closed. The complement is path-connected. There is no path in the complement to $(0,1/2)$.


While this set is not simply connected, that's not important. The intersection with the upper half plane is simply connected and has the same property. Your argument using the Riemann mapping theorem assumed that there is a continuous extension of the Riemann map to the boundary, but that might not be possible when the complement is not locally connected.


