Perhaps the theorem at the common intersection of the various ambiguous interpretations of the inquiry at hand is the following:
If the planar open set U contains a continuum K, and if the planar complement of K is connected, then there exists a connected and simply connected open set V such that K is a subset of V, and V is a subset of U.
This follows, for example, from the fact that K is cellular, the nested intersection of closed topological planar disks. Surround K by a slightly larger disk D, so that D is contained in U, and the interior of D is the coveted open set V.
For a barehanded argument of the cellularity claim, prove by induction that if C is a planar simple closed curve, and if C is the cancatanation of vertical and horizontal segments of length 1, then C bounds a topological disk.
To obtain the mentioned disk D, tile the plane by very small squares, and extract a curve C from the boundaries of the tiled squares.
Of course the argument sketched above does not use any heavy artillery such as the Jordan curve theorem or the Schoenflies theorem.