Question: in Section 1.3 "Connected and compact sets" from the book "Lecture notes on elementary Topology and Geometry" by Singer & Thorpe, Theorem 2 states the following:

*Theorem*: Let *S* be a topological space (with **U** as the topology), and let $T\_{0}$ and {$T\_{w}$} with ${w \in W}$ connected subsets of *S* (that is, connected with the relative topology or subspace topology). Assume $T\_{0} \cap T\_{w} \neq \varnothing$ for each ${w \in W}$. Then, $T\_{0} \cup (\cup\_{w\in W} T\_{w})$ is connected.

This can be used to prove that $\mathbb{R^n}$ is connected, because $\mathbb{R^n}$= {0} $\cup (\cup\_{w\in W} T\_{w})$ where $T\_{w}$ are all the lines through $0$ and $W$ can be the unit sphere in $\mathbb{R^n}$.

So I have two questions. 

1. If $n=2$ how can I use this theorem to prove that $\mathbb{R^2}$-{0} is connected?
2. Can I use the answer to 1. to prove that $\mathbb{R^n}$-{0} is connected?

I ask firstly for $n=2$ because that is what I used to understand the proof that $\mathbb{R^n}$ is connected, but I really want to undestand the general case. Thanks in advance.

Answer: 
Pick any point p /= 0, the origin.  

For q /= 0, draw a line from p to q.  

If pq passes through 0, then draw a broken line  

from p to q avoiding 0.  

All these lines are connected, cover $R^n$ - {0}  

and have p in common. Thus $R^n$ - {0} is connected,  

path connected in fact. 

Exercise. Let A be a countable set, n > 1.  

Show $R^n$ - A is connected, even path connected.
Let $T\_0=\Bbb R\times \{1\}$ and let $A=\{T\_w\}\_{w\in W}$ be the set of all the lines in $\Bbb R^2\setminus \{(0,0)\}$ that intersect $T\_0.$ Every $(x,y)\in \Bbb R^2\setminus \{0\}$ belongs to a member of $A:$ If $x\ne 0$ then $(x,y)\in \{x\}\times \Bbb R\in A.$ If $y\ne 0$ the line $\{(u,u+y):u\in \Bbb R\}$ thru $(0,y)$ and $(1,1+y))$ belongs to $A.$

Similarly for $\Bbb R^n \setminus \{0\}^n$ when $n\geq 3.$ Every point in $\Bbb R^n\setminus \{0\}^n$ lies on a line in $\Bbb R^n\setminus \{0\}^n $that intersects the line $\Bbb R\times \{1\}^{n-1}.$