Prove that removing an open neighbourhood $V$ of a knot $K$ in $S^3$ does not affect the connectedness of the complement, $S^3 \backslash V$

Question

I saw that answer: $W$ be an $m$-dimensional linear subspace of $\mathbb R^n$ such that $m\le n-2$ , then is it true that $\mathbb R^n \setminus W$ is connected?, but I am not sure how can I apply it to $S^3$ or rather: if I can apply this. Does it somehow follow from the one-point compactification of $\mathbb{R}^3$?