Prove that $\mathbb{R}^n\setminus \{0\} $ is connected for $n > 1$ 
Prove that $\mathbb{R}^n\setminus \{0\} $ is connected for $n > 1$.

I don't understand where to start proving this since
$$\mathbb{R}^n \setminus \{0\} = (-\infty,0)^n \cup (0, \infty)^n $$
Which is the union of two disjoint nonempty open sets, so it can't be connected. Obviously I won't be told to prove something is true that isn't so I know I'm missing something.
We have proven that $\mathbb{R}^n$ is connected using this theorem:
Let S be a topological space, and let $T_0$ and $\{T_w\}_{w\in W} $ be connected subsets of S. Assume that $T_0 \cap T_w \neq \emptyset $ for each w. Then $T_0 \cup \left( \cup_{w \in W} T_w \right)$ is connected.
Using the first connected set {0} and the indexed ones as lines that go through the point {0} indexed by the unit sphere. 
I was hoping to do something similar with this problem, but I can't see a way to do that. Help would be appreciated. And apologies for any Latex mistakes. I'll try to fix them but I'm on vacation and only have my phone currently.
 A: Your "formula" for $\mathbb{R}^n\backslash \{0\}$ is wrong. Draw $\mathbb{R}^2$ and you will understand why.
If you want to avoid usage of path-connectedness, do the following. For $n>1$, take $A,B \subset \mathbb{R}^n \backslash \{0\}$ given by $A:=\{x \in \mathbb{R}^n \mid x_n > 0\}$ and $B:=\{x \in \mathbb{R}^n \mid x_n<0\}$. $A$ and $B$ are clearly connected (both in fact homeomorphic to $\mathbb{R}^n$). Since the closure of connected sets is connected, $\overline{A}$ and $\overline{B}$ are connected. But $\overline{A}$ and $\overline{B}$ have points in common. Therefore, their union is connected. But their union is the entire $\mathbb{R}^n\backslash \{0\}$ .
Fun exercise: Where does this fail for $\mathbb{R}$?

 "But $\overline{A}$ and $\overline{B}$ have points in common" - They don't, for $n=1$.

A: Hint:  It's usually way easier to show path connected than connected.  Path connected implies connected, it's stronger.  So take two arbitrary points and show you can connect them with a path.   There are two cases:  The points are not on the same line as the origin on opposite sides (The easy case), and they are:  (The almost as easy case)
Can you manage?
A: There is a path from any point to the unit disc,
$r(t)= ((1-t)+{t \over \|x\|}) x$
Now all that remains is to show that the unit disk is connected. Pick $u_1,u_2$ on the unit disk. Let $v_1=u_1$ and
$v_2$ on the unit disc in the span of $u_1,u_2$ such that $v_1 \bot v_2$.
Now consider the path $\rho(\theta) = (\cos \theta) v_1 + (\sin \theta) v_2$,
and show that $\rho(\theta)$ is on the unit disk for all $\theta$ and that there is some $\theta'$ such that $\rho(\theta') = u_2$.
Here is an explicit construction: Let
$v_2 = { u_2 -\langle u_1, u_2 \rangle u_1 \over \| u_2 -\langle u_1, u_2 \rangle u_1  \| }$. Note that $v_2$ is on the unit disk and we have
$u_2 = \| u_2 -\langle u_1, u_2 \rangle u_1  \| v_2 + \langle u_1, u_2 \rangle u_1$. Now check that $\| u_2 -\langle u_1, u_2 \rangle u_1  \|^2 + (\langle u_1, u_2 \rangle u_1)^2 = 1$, hence there is some $\theta'$ such that
$\sin \theta' = \langle u_1, u_2 \rangle u_1$ and
$\cos \theta' = \| u_2 -\langle u_1, u_2 \rangle u_1  \|$, from which we have
$\rho(\theta') = u_2$. Hence the map $\rho:[0,\theta'] \to S^{n-1}$ is a path connecting $u_1,u_2$ in $S^{n-1}$.
A: $n$-dimensional polar coordinates give rise to a continuous surjection from the cartesian product of connected intervals to $\mathbb R^n \setminus \{0\}$.
