The complement of a polygonal arc in ${\mathbb{R}}^2$ is polygonally connected

Complete $A$ to a polygon $P=A\cup B$, where $B$ is a polygonal arc with the same endpoints as $A$, has no other point in common with $A$, and avoids $x,y$. To construct $B$, take a sufficiently thin polygonal regular neighbourhood of the finite arc $A$: disjoint small vertex neighbourhoods joined by narrow strips along the edge interiors form a polygonal disk, and either boundary chain, capped to the two endpoints, supplies $B$. General position permits the finitely many boundary vertices to avoid $x,y$.

By [L1], $P$ has regions $U$ and $V$. If $x,y$ lie in the same region, polygonal connectedness of open components joins them there. If they lie in opposite regions, choose a point $z$ in the relative interior of $B$. A sufficiently short segment transverse to $B$ at $z$ has one endpoint $z_U$ in $U$ and the other $z_V$ in $V$. Join $x$ to the endpoint on its side and $y$ to the other by polygonal paths within those regions, then concatenate those paths with $z_U-z-z_V$.

The paths in step 2.1 avoid all of $P$ except possibly at $z\in B$, and $B\cap A$ consists only of the two common endpoints, so the concatenated path avoids $A$. Since $x,y$ were arbitrary, ${\mathbb{R}}^2\setminus A$ is polygonally connected.