On a special case of Alexander duality  Let $S^n$ be the $n$-dimensional sphere and let $K\subseteq S^n$ be a compact, locally contractible subspace of real codimension $\geq 2$. Applying Alexander duality we find that
$$
\tilde{H}_{i}(S^n-K)\simeq \tilde{H}^{n-i-1}(K)
$$
where $\tilde{H}$ denotes the reduced homology (cohomology) with coefficients in  $\mathbf{Z}$ and $0\leq i\leq n-1$. In particular, taking $i=0$ we find that 
$$
\tilde{H}_{0}(S^n-K)\simeq \tilde{H}^{n-1}(K)=0.
$$
Thus $S^n-K$ is connected and therefore path connected (since $S^n-K$ is a euclidian open set and therefore locally path connected).
Q: Is there a simple (low-tech and/or geometrical proof) that $S^n-K$ is path connected ?
 A: Since you're asking for a geometric proof, let's just work with manifolds. Then the general claim is that if $N\subset M$ is a codimension 2 submanifold of a connected manifold, then $M\setminus N$ is connected. Iterating the claim, we can reduce to the case when $N$ is also connected.
By the tubular neighborhood theorem, we reduce$^{\dagger}$ to showing that the (total space of the) normal bundle of $N$ in $M$ minus the zero section is connected. I.e., we reduce to the case of showing that a vector bundle of rank $\geq 2$ minus its zero section is connected. 
This is clearly true, but I'll include one possible proof for completeness' sake. Take two points in the vector bundle $\mathcal{E}$ over $N$ which don't lie in the zero section. Take a smooth simple path in $N$ between their projections, and then take a tubular neighborhood of this path. This neighborhood is contractible and $\mathcal{E}$ is a vector bundle over it, so it's a trivial vector bundle. Therefore, we reduce to showing that $\mathbb{R}^n$ minus a codimension 2 linear subspace is connected, which is unobjectionable.
$^{\dagger}$ Here's how you do the reduction. Take two points in $M\setminus N$ and a path between them in $M$. If this path doesn't go through $N$, we're certainly in good shape. Otherwise, choose two points on the path in the neighborhood, one "before" the path crosses through $N$, one "after" (i.e., after it has finished passing through $N$). Since the tubular neighborhood is diffeomorphic to the normal bundle of $N$ in $M$, we can choose a path in this tubular neighborhood which doesn't cross $N$, and modifying our original path by this procedure, we get the result.
A: 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).
