Proving that $\{ x \in \mathbb{R}^n : |x| = 1, x\geq 0\}$ is homeomorphic to $\overline{B_1(0)} \subset \mathbb{R}^{n-1}$ I want to show that  $A= \{ x \in \mathbb{R}^n : |x| = 1, x\geq 0\}$ is homeomorphic to $\overline{B_1(0)} \subset \mathbb{R}^{n-1}$, where $\overline{B_1(0)}$ is the closed unit ball. I was thinking about the following simple proof:
We have $A \subset S^{n-1}$ where $S^{n-1}$ is the unit sphere in $\mathbb{R}^n$. Using higher dimensional spherical coordinates it follows that $A$ is homeomorphic to
$$
[0,\pi/2]^{n-1} \subset \mathbb{R}^{n-1}.
$$
But this is a square which is homeomorphic to $B_1(0) \subset \mathbb{R}^{n-1}$.
Is this a rigorous approach?
 A: As you clarified in a comment, by $x = (x_1,\dots,x_n) \ge 0$ you mean that all $x_i \ge 0$. Moreover, I guess that $\lvert - \rvert$ denotes the Euclidean norm. In the sequel let $n \ge 2$.
The projection $p_n : \mathbb R ^n \to \mathbb R ^{n-1}, p_n(x_1,\dots,x_n) = (x_1,\dots,x_{n-1})$, maps $A$ homeomorphically onto $B^{n-1} = \{y \in \mathbb R ^{n-1} \mid \lvert y \rvert \le 1, y \ge 0 \}$.
In fact, $p(A) \subset B^{n-1}$ because $\lvert (x_1,\dots,x_{n-1}) \rvert \le \lvert (x_1,\dots,x_n) \rvert = 1$ and all $x_i \ge 0$. Let $p_A : A \to B^{n-1}$ denote the restriction of $p_n$. Moreover, define $f : B^{n-1} \to A, f(y) = (y, \sqrt{1 - \lvert y \rvert^2})$. Note that $\lvert f(y) \rvert = 1$ and $f(y) \ge 0$. It is easy to see that $p_A \circ f = id_{B^{n-1}}$ and $f \circ p_A = id_A$ which means that $p_A$ is a homeomorphism whose inverse is $f$.
Let $m = n-1$. We prove by induction on $m$ that there exists a homeomorphism $h_m : B^m \to D^m = \overline{B_1(0)}$.
For $m=1$ we have $B^1 = [0,1]$ and $D^1 = [-1,1]$. $h_1$ is given by $h_1(x) = 2x - 1$.
Given $h_{m-1}$, we construct $h_m$ as follows.
First observe that $B^m \subset B^{m-1} \times [0,1]$. For $(x,x_m) \in B^m$ define
$$h'_m(x,x_m) =
\begin{cases}
(\sqrt{1-x_m^2}h_{m-1}(\frac{1}{\sqrt{1-x_m^2}}x) ,x_m) & x_m < 1 \\
(0,1) & x_m = 1
\end{cases}
$$
Note that this is well-defined because $(x,x_m) \in B^m$ means $x \ge 0$ and $\lvert x \rvert \le \sqrt{1-x_m^2}$. Hence $\frac{1}{\sqrt{1-x_m^2}}x \in B^{m-1}$. Moreover $\left\lvert \sqrt{1-x_m^2}h_{m-1}(\frac{1}{\sqrt{1-x_m^2}}x) \right\rvert \le \sqrt{1-x_m^2}$ which means $h'_m(x,x_m) \in D^m_+$ = closed upper half of the unit ball $D^m$. Also note that if $x_m =1$, then necessarily $x = 0$.
$h'_m$ is obviously continuous at all $(x,x_m)$ with $x_m < 1$. It is also continuous at $(0,1)$. Let $(\xi_n,t_n) \to (0,1)$. Then $h'_m(\xi_n,t_n) \to
 (0,1)$ because $\sqrt{1-t_n^2} \to 0$ and $\left\lvert \sqrt{1-t_n^2} h_{m-1}((\frac{1}{\sqrt{1-t_n^2}}\xi_n) \right\rvert \le \sqrt{1-t_n^2}$.
Geometrically $h'_m$ does the following: The slice $B^m \cap (\mathbb R ^{m-1} \times \{ x_m \})$ is linearly expanded to $B^{m-1} \times \{ x_m \}$, which is mapped homeomorphically onto $D ^{m-1} \times \{ x_m \}$, which is linearly compressed to $D^m \cap (\mathbb R ^{m-1} \times \{ x_m \})$.
$h'_m : B^m \to D^m_+$ is a homeomorphism. In fact, its inverse is given by
$$k(x,x_m) = 
\begin{cases}
(\sqrt{1-x_m^2}h^{-1}_{m-1}(\frac{1}{\sqrt{1-x_m^2}}x) ,x_m) & x_m < 1 \\
(0,1) & x_m = 1
\end{cases}
$$
The final step is to construct a homeomorphism $g_m : D^m_+ \to D^m$. Then $h_m = g_m \circ h'_m$ will do. We can take
$$g_m(x,x_m) = (x, 2x_m - \sqrt{1 - \lvert x \rvert^2}) .$$
The necessary checkings are left to you. Geometrically $g_m$ maps the line segment $\{x \} \times [0,\sqrt{1 - \lvert x \rvert^2}]$ linearly onto $\{x \} \times [- \sqrt{1 - \lvert x \rvert^2} ,\sqrt{1 - \lvert x \rvert^2}]$. 
A: Here is a fact that you might want to prove: the $n$-sphere minus one point is homeomorphic to $\mathbb R^n.$ The $n$-sphere is the set of points in $\mathbb R^{n+1}$ with radius $1$.
For example, a sphere with a point plucked out is the plane.
It is usually even mentioned in introductory courses. Google 'stereographic projection' if you want more details.
If you understand this well enough, you can prove the statement.
A: A sketch of a possible approach: first we show that the boundaries of cubes are homeomorphic to boundaries of balls. This follows because norms on $\mathbb R^n$ are equivalent. Or you can just find maps that work: $x\mapsto \frac{x}{\|x\|_\infty}$ is bijective continuous and maps the boundary of the ball to the boundary of the cube, with continuous inverse $x\mapsto \frac{x}{\|x\|_2}$ 
So, now we want to prove that the closed $n-1$ dimensional ball is homeomorphic to the boundary of an $n$ dimensional ball. We prove instead that the upper hemisphere of the $n-1$-ball is homeomorphic to boundary of the  upper hemisphere of the $n$-ball. A similar result will hold on the lower hemispheres, and the proof will then follow by the pasting lemma. 
A homeormorphism is provided by the map 
$\phi:(x_1,x_2,\cdots, x_{n-1}) \mapsto (x_1,x_2,\cdots, x_{n-1},b)$ 
where $b$ is the $unique$ number such that $\sum x_i^2+b^2=1,$ 
