I am hesitant to use the Jordan curve theorem here but perhaps it is possible. Instead what I would use is the important fact that the space $Emb(D^n , M)$ is equivalent to $Fr(M)$, the total space of the fiber bundle over $M$ where the fiber over $x$ is the space of bases of the tangent bundle. Here $M$ is n-dimensional. This equivalence just comes from pushing forward the standard basis along the derivative.
For a nonorientable connected manifold, this space is path connected because the base is path connected and there exists a path in the total space that connects any two points in a specific fiber (this is equivalent to being nonorientable).
From this we conclude that any two embeddings of the disk are isotopic. A very nice result called isotopy extension then implies that there is a diffeomorphism of the manifold taking one of these disks to the other. This implies that as pairs (the manifold and the image of the embedded disk), they are exactly the same topologically (as well as smoothly). This implies that removing the interior gives homeomorphic (as well as diffeomorphic) manifolds.