The proof that $S^n \setminus K$ is path connected follows from directly general position, if $K$ is, say a finite simplicial complex of codimension two (I don't know what "codimension two" means in your general formulation). 

If we take two points $x,y$ not in $K$, and we take a generic path in $S^n$ connecting them, then general position taken with respect to each simplex of $K$ guarantees that
a generic path won't intersect $K$.