The proof that $S^n \setminus K$ is path connected follows directly from 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 $K$.

The proof that $S^n \setminus K$ is path connected follows directly from 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 $K$.

Addendum: a more general fact is true: if $K \subset S^n$ is a simplicial complex of codimension $m$, then general position implies that the complement $S^n \setminus K$ is $(m-2)$-connected (the proof is similar).

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 $K$.

The proof that $S^n \setminus K$ is path connected follows directly from 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 $K$.