I am only releasing part of the proof.
Doesn't this prove that "otherwise" case is completely wrong? I am assuming "otherwise" refers to $n < 2$? Because he just showed that $H^0(\Bbb R - \{ p \}) \simeq \Bbb R^2 $, not isomorphic to $0$. I should mention this exercpt is is setup for the base case of the eventual induction on $k$. The proof becomes long, so I didn't post it here, nor it is relevant.
For some reason, the problem computes the cohomology for $n\geq 2$, but the author starts the base case at $n=1$! It's not the "otherwise" that's the problem. This can't possibly refer to $n<2$: that's simply not how scope of variables works in any kind of mathematical writing. It must refer, instead, to $d\notin\{0,n-1\}$, since those are the two cases covered by the other lines. So there's a contradiction if you try to read this with $n=1$, because we're given two different suggestions for $d=0$, namely, $\mathbb{R}^2$ from the first line and $\mathbb{R}$ from the second. The proposition should read $$H^d(\mathbb{R}^n-\{p_1...p_k\})=\left\{\begin{matrix} \mathbb{R}^k, d=n-1\\\mathbb{R}, d=0\text{ and }n> 1\\ 0,\text{ otherwise, i.e. }0<d<n-1\text{ or }d\geq n \end{matrix}\right.$$