$\mathbb{R}^{p}\backslash \{0 \}$ is connected for $p \geq 2$ I have to show that $\mathbb{R}^{p}\backslash \{0 \}$ is connected for $p \geq 2$. Is it possible to show this using the property debated on in this article: Union of connected subsets is connected if intersection is nonempty?
 A: $\Bbb R^p\setminus\{0\}$ is the union of the $2p$ open halfspaces $x_1>0$, $x_1<0$, $x_2>0$, $x_2<0$, $\ldots$, $x_p>0$, $x_p<0$.
Halfspaces are convex and hence connected.
EDIT: Certainly, the intersection of all the halfspaces is empty. But they still make the job:
Indeed: let $H_j^+=\{x_j>0\}$ and $H_j^-=\{x_j<0\}$, for $j\in\{1,\ldots,p\}$. Define:
$$A_1=H_1^+,\; A_{j+1}=A_j\cup H_{j+1}^+\text{, for }2\le j\le p-1$$ $$A_{p+1}=A_p\cup H_1^-,\; A_{p+j+1}=A_{p+j}\cup H_{p+j+1}^-\text{, for }2\le j\le p-1$$
Then each $A_j$ is connected because is the non-disjoint union of two connected sets. And $A_{2p}=\Bbb R^p\setminus\{0\}$.
Note that this reasoning does not work (as it should be) if $p=1$ because $A_1\cap A_2=\emptyset$. 
A: As observed, you can cover $\Bbb R^p\setminus\{0\}$ with half-spaces in the natural way.
However, said covering does not satisfy the hypothesis of the theorem you've linked: there are in fact $p$ couples of disjoint subsets!
To proceed with that idea, I suggest a slight variation of the theorem you've linked (which you can prove as an exercise):

Let $X$ be a topological space and let $\mathfrak U$ be a covering of $X$ in non-empty connected subspaces. If for any $A,B\in \mathfrak U$ such that $A\cap B=\emptyset$ there is $C\in \mathfrak U$ such that $C\cap A\ne\emptyset$ and $C\cap B\ne\emptyset$, then $X$ is connected.

