I must prove that, being $S$ a differentiable surface of dimension $d$, which is a closed subset of $\mathbb{R}^k$, where $k\geq d+2$, one have $\mathbb{R}^k-S$ connected.
This is trivial for $d=1$, because so $S$ is (a subset of) one line in a space at least $\mathbb{R}^3$, so one can deviate of this line and obsviouly the complementary is connected.
However, I have doubts to formalize this argument in such a way as to make it extendable.
I'd appreciate some ideas. However, for ethical reasons, I ask you to be kind enough not to post detailed answers.
For any compact, locally contractible, nonempty, proper subspace of $K$ of $n$ dimensional sphere $\mathbb{S}^n$, Alexander Duality says: $\tilde{H}_i(\mathbb{S}^n-K;\mathbb{Z})\simeq\tilde{H}^{n-i-1}(K;\mathbb{Z})$ for all $i$.
Thinking $\mathbb{S}^n$ as one point compactification of $\mathbb{R}^n$ and applying Alexander Duality to the cases when $K$ is compact $\mathbb{R}^n$ and non-compact in $\mathbb{R}^n$ gives the result since $\tilde{H}_0(\mathbb{S}^n-\overline{K};\mathbb{Z})\simeq 0$ where $\overline{K}$ is the closure of $K$ in $\mathbb{S}^n$. This means we have path-connectedness for $\mathbb{S}^n-\overline{K}$. Hence connectedness for $\mathbb{S}^n-\overline{K}$. Now you need to argue removing one point back does not change the connectedness.