On homeomorphism between complements of closed disc and $[0,1]^3$ $\mathbf {The \ Problem \ is}:$  Show that the spaces $\mathbb R^3 - D^3$ and $\mathbb R^3 - I^3$ are homeomorphic , where $I = [0,1]$, $D$ denotes the closed unit ball and $\mathbb R^3$ is under standard topology .
$\mathbf {My \ approach} :$ Actually, I tried to show that $\mathbb R^3 - I^3 \cong \mathbb R^3 -\{(0,0,0)\} =\{(r,\theta,\phi) | 0\lt r\lt \infty ; 0\leq \theta,\phi \lt 2π\}$
by spherical co-ordinates as the right side is equivalent to $\mathbb R^3 - D^3 = \{(r,\theta,\phi) | r\gt 1 ; 0\leq \theta,\phi \lt 2π\}$ by component-wise homeos, 
 $(0,\infty) \mapsto (1,\infty)$ and $[0,2π) \mapsto [0,2π) .$
Now, I think for any vector $p \in \mathbb R^3 - \{(0,0,0)\}$, if we  draw a line joining $p$ with $\{(0,0,0)\}$ and strech it along that line, in a fixed direction ,in such a way to get it out of $[-1,1]^3$ as $[0,1]^3 \cong [-1,1]^3$ (by component-wise linear stretching) . 
But I am lacking to rigourously produce a continuous bijection .
A small hint is warmly appreciated. 
 A: As you noticed, it suffices to show that both $\mathbb R^3 \setminus D^3$ and $\mathbb R^3 \setminus I^3$ are homeomorphic to $\mathbb R^3 \setminus \{(0,0,0)\}$.
A homeomorphism $\phi : \mathbb R^3 \setminus D^3 \to \mathbb R^3 \setminus \{(0,0,0)\}$ is given by $\phi(x) = \dfrac{\lVert x \rVert - 1}{\lVert x \rVert }x$. Its inverse is  $y \mapsto \dfrac{\lVert y \rVert + 1}{\lVert y \rVert }y$.
Next consider the homeomorphism $h : \mathbb R^3 \to \mathbb R^3, h(x) = 2x - (1,1,1)$. We have $h(I^3) = [-1,1]^3 = J^3$, thus $h$ restricts to a homeomorphism $R^3 \setminus I^3 \to R^3 \setminus J^3$. 
The maximum norm on $\mathbb R^3$ is defined by $\lVert (x_1,x_2,x_3) \rVert_\infty = \max(\lvert x_1 \rvert, \lvert x_2 \rvert, \lvert x_3 \rvert)$. We have $J^3 = \{ x \in \mathbb R^3 \mid  \lVert x \rVert_\infty \le 1 \}$. Let us verify that the norm $\lVert - \rVert_\infty : \mathbb R^3 \to \mathbb R$ is a continuous function. For $x = (x_1,x_2,x_3)$ we have $\lvert x_i \rvert = \sqrt{x_i^2} \le \sqrt{x_1^2 + x_2^2 + x_3^2} = \lVert (x_1,x_2,x_3) \rVert$ and we see that $\lVert x \rVert_\infty \le \lVert x \rVert$. Since $\lvert \lVert x \rVert_\infty - \lVert x' \rVert_\infty \rvert \le \lVert x - x' \rVert_\infty$ via the triangle inequality, we conclude that $\lvert \lVert x \rVert_\infty - \lVert x' \rVert_\infty \rvert \le \lVert x - x' \rVert$ which proves the continuity of the maximum norm.
Now a homeomorphism $\psi : \mathbb R^3 \setminus J^3 \to \mathbb R^3 \setminus \{(0,0,0)\}$ is given by $\psi(x) = \dfrac{\lVert x \rVert_\infty - 1}{\lVert x \rVert_\infty}x$.
Remark:
A homeomorphism $H : \mathbb R^3 \setminus D^3 \to \mathbb R^3 \setminus J^3$ can also be given directly by
$$H(x) = \dfrac{\lVert x \rVert}{\lVert x \rVert_\infty}x . $$
You see that $\lVert H(x) \rVert_\infty = \lVert x \rVert$.
