Homeomorphism between $\mathbb{R}^{n} - \{0\}$ and $S^{n-1} \times \mathbb{R}$ I am trying to make a homeomorphism between $\mathbb{R}^{n}-\{0\}$ and $S^{n-1} \times \mathbb{R}$. 
My attempt is to use the graphic and the domain of continuos functions are homeomorphic. 
Then define $f:\mathbb{R}^{n}- \{0\} \to \mathbb{R}$,  $f(x_1,\dots,x_{n})= \sqrt{1 - x_1^2 + \cdots x_{n}^2 }.$ 
So I can conclude that the graphics $G = \{(x_1,x_2,\dots,x_{n},f(x)) \in \mathbb{R}^{n+1}: x_1^2 + \cdots x_{n}^2 + f(x)^2 =1\}$ is homeomorphic to $\mathbb{R}^{n}-\{0\}.$
But, I see that $G = S^n$, how can I proof that $S^n $ is homeomorphic to $S^{n-1} \times \mathbb{R}??$

You can't prove that last thing, because it's not true. 
But the homeomorphism you seek looks like this:
$$
H: S^{n-1} \times \Bbb R : ((x_1, \ldots, x_n); t) \mapsto 
f(t)(x_1, \ldots, x_n)
$$
where $f: R \to R^{+}$ is any homeomorphism from $R$ to $R^{+}$, such as $t \mapsto \exp(t)$. 
So at least one solution is
$$
H(x_1, \ldots, x_n, t) \mapsto (e^t x_1, \ldots, e^t x_n).
$$
Of course, you still need to prove that it's a homeomorphism, but that may be easier for you. 
