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.