How do I prove that this function between an octahedron and a 3 dimensional disk is a homeomorphism? Define $f:R\rightarrow D^3$ such that for $(x,y,z)\neq (0,0,0)$, $f(x,y,z)=\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}(x,y,z)$ and $f(0,0,0)=(0,0,0)$.
Specifically, how do I prove that it is continuous? If you know of more than one way to prove this, could you please share? Any help is appreciated.
Edit: I need to show that $R=\{(x, y, z) \in \mathbb{R}^3| |x|+|y|+|z|\leq 1\}$ is homeomorphic with the 3-dimensional disk $D^3=\{(x, y, z)\in \mathbb{R}^3|x^2+y^2+z^2\leq 1\}$ 
$f$ is a map from $R$ to $D^3$.
 A: This is similar as On homeomorphism between complements of closed disc and $[0,1]^3$.
On $\mathbb R^3$ we have the Euclidean norm $\lVert (x_1,x_2,x_3) \rVert = \sqrt{x_1^2 + x_2^2 + x_3^2}$ which induces the Euclidean metric $d(x,y) = \lVert x - y \rVert$. The continuity of maps between subsets of $\mathbb R^3$ is usually defined via the Euclidean metric (alternatively it can be defined via the topology of $\mathbb R^3$ generated by $d$ and the subspace topologies on the subsets).
Another norm on $\mathbb R^3$ is defined by $\lVert (x_1,x_2,x_3) \rVert_1 = \lvert x_1 \rvert + \lvert x_2 \rvert + \lvert x_3 \rvert$. You should do the straightforward verification that it is a fact a norm. We have $R = \{ x \in \mathbb R^3 \mid \lVert x \rVert_1 \le 1 \}$.
Let us verify that the norm $\lVert - \rVert_1 : \mathbb R^3 \to \mathbb R$ is a continuous function.  For $x = (x_1,x_2,x_3)$ we have $\lvert x_i \rvert = \sqrt{x_i^2} \le \sqrt{x_1^2 + x_2^2 + x_3^2} = \lVert (x_1,x_2,x_3) \rVert$ which implies $\lVert x \rVert_1 \le 3\lVert x \rVert$. Since $\lvert \lVert x \rVert_1 - \lVert x' \rVert_1 \rvert \le \lVert x - x' \rVert_1$ via the triangle inequality, we conclude that $\lvert \lVert x \rVert_1 - \lVert x' \rVert_1 \rvert \le 3\lVert x - x' \rVert$ which proves the continuity of the norm.
Moreover we have $\lVert (x_1,x_2,x_3) \rVert^2 = x_1^2 + x_2^2 + x_3^2 \le (\lvert x_1 \rvert + \lvert x_2 \rvert + \lvert x_3 \rvert)^2 = \lVert (x_1,x_2,x_3) \rVert_1^2$, i.e. $\lVert x \rVert \le \lVert x \rVert_1$. There are $x$ with $\lVert x \rVert < \lVert x \rVert_1$, for example $x = (1,1,1)$.
Now define
$$\phi : D^3  \to R, \phi(x) = \begin{cases}
0 & x = 0 \\
\dfrac{\lVert x \rVert}{\lVert x \rVert_1}x & x \ne 0
\end{cases}$$
You see that $\lVert \phi(x) \rVert_1 = \lVert x \rVert$, i.e. $\phi$ is well-defined. This shows also that $\phi$ is continuous: This is obvious in all points $\xi \ne 0$, and continuity in $0$ follows from $\lVert \phi(x) - \phi(0) \lVert = \lVert \phi(x)  \lVert  \le \lVert \phi(x) \lVert_1 = \lVert x \lVert = \lVert x - 0 \lVert$.
Similarly
$$\psi : R  \to D^3, \psi(x) = \begin{cases}
0 & x = 0 \\
\dfrac{\lVert x \rVert_1}{\lVert x \rVert}x & x \ne 0
\end{cases}$$
is a well-defined continuous map. We have $\psi \circ \phi = id$ and $\phi \circ \psi = id$ which shows that $\phi, \psi$ are inverse homeomorphisms.
Note that your map $f : R \to D^3$ is not surjective, thus not a homeomorphism. For example, take $\xi = \frac{1}{\sqrt{3}}(1,1,1)$. Then $\lVert \xi \rVert = 1$. Assume there is $x \in R$ such that $f(x) = \xi$. Then $\dfrac{\lVert x \rVert}{\lVert x \rVert_1}x = \xi$ which implies $\lVert x \rVert = \lVert \xi \rVert_1 = 3\frac{1}{\sqrt{3}} = \sqrt{3} > 1$. We conclude $\lVert x \rVert_1 > 1$ which is impossible for $x \in R$.
