Show that the closed unit-ball in $\mathbb{R}^2$ is homeomorphic to the unit-2 sphere in $\mathbb{R}^3$ This problem is taken from Topology: A First Course by Munkres

Let $X$ be the closed unit ball $$\{ \langle x,  y\rangle \mid x^2 + y^2 \le 1\}$$ in $\mathbb{R}^2$, and let $X^{\ast}$ be the partition of $X$ consisting of all the one-point sets $\{ \langle x,  y\rangle\}$ for which $x^2 + y^2 < 1$, along with the set $S^1= \{ \langle x,  y\rangle \mid x^2 + y^2 = 1 \}$.  One can show that $X^{\ast}$ is homeomorphic with the subspace of $\mathbb{R}^3$ called the unit 2-sphere, defined by $$S^2 = \{\langle x,  y, z\rangle \mid x^2 + y^2 + z^2 =1 \}.$$

My Attempted Proof
$(X^*, \mathcal{T})$ is the quotient space of $X$ where $\mathcal{T}$ is the topology induces by the quotient map $p : X \to X^*$. $$\mathcal{T} = \left\{U \subset X^* \ \middle| \ p^{-1}(U) = X \cap W \text{ for $W$ open in $\mathbb{R}^2$}\right\}$$
Since $S^2 \subset \mathbb{R}^3$, the topology $\mathcal{M}$ on $S^2$ is given by $$\mathcal{M} = \left\{S^2 \cap V \ | \ V \text{ open in $\mathbb{R}^3$}\right\}$$
Now define $h : X^* \to S^2$ by $$h\left(\langle x, y \rangle\right) = \begin{cases} \langle x, y, \sqrt{1-x^2 + y^2} \rangle \ \ \  \text{ if $x^2 + y^2 < 1$} \\
\langle x, y, 0 \rangle \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \text{ if $x^2 + y^2 = 1$} \\
\end{cases}$$
We now show that $h$ is a homeomorphism. Pick $U \in \mathcal{T}$, we first show that $h(U) \in \mathcal{M}$. $U$ is of the form $U = \{ \langle x, y \rangle \ | \ x^2 + y^2 = 1 \text{ or } x^2 + y^2 < 1 \}$, and by definition of $h$ we have $h(U) = \{\langle x, y , z\rangle \ | \ x^2 + y^2 +z^2 = 1 \} \subset \mathbb{R}^3$. Fix $\epsilon > 0$ and choose $V = (-1 - \epsilon, 1 + \epsilon) \times (-1 - \epsilon, 1 + \epsilon) \times (-1 - \epsilon, 1 + \epsilon)$. Then $V$ is open in $\mathbb{R}^3$ and we have $h(U) \cap V = h(U) \in \mathcal{M}$. Thus $U \in \mathcal{T} \implies h(U) \in \mathcal{M}$
We now have to prove that $h(U) \in \mathcal{M} \implies U \in \mathcal{T}$. 

But that is where I got stuck. I'm not show how to show $U \in \mathcal{T}$ if we pick $h(U) \in \mathcal{M}$.
Am I on the correct track, are there any errors so far in my proof, is it non-rigorous. If so, please let me know. If not then how can I go about proving the reverse implication to show that $h$ is a homeomorphism?
 A: OK, so the question is clearer if you write it like this:
There's a relation on a $n$-dimensional closed unit ball $D^n$:
$$x\sim y\mbox{ if and only if }x=y\mbox{ or }\lVert x\rVert =\lVert y\rVert =1$$
Prove that $D^n\ /\sim$ is homeomorphic to $S^n$.
First of all note that your function $h$ is not a homeomorphism because it is not "onto" (your last coordinate is always nonnegative).

Proof. Start by looking the other way around. Define
$$\theta:\mathbb{R}^{n}\to S^n$$
$$\theta(x_1, \ldots, x_n)=\bigg(\frac{S-1}{S+1}, \frac{2x_1}{S+1},\cdots,\frac{2x_n}{S+1}\bigg)$$
where $S=\sum x_i^2$. This map is also known as the stereographic projection (or more accurately its inverse). Note that the image consists of all points on the sphere except for $(1,0,\ldots, 0)$. Also $\theta$ is one-to-one.
Now define a map from open disk to $\mathbb{R}^n$:
$$g:B^n\to\mathbb{R}^n$$
$$g(x)=\begin{cases}
\tan\bigg(\frac{\pi}{2}\lVert x\rVert\bigg)\frac{x}{\lVert x\rVert}\mbox{ if }x\neq 0 \\
0\mbox{ otherwise }
\end{cases}$$
This is a homeomorphism, an inverse given by scalar multiplication by $\arctan$.
Finally we are ready to define our function:
$$f:D^{n}\to S^n$$
$$f(x)=\begin{cases}
\theta\big(g(x)\big)\mbox{ if } \lVert x\rVert < 1 \\
(1,0,\ldots,0)\mbox{ otherwise}
\end{cases}$$
You can easily check that it is continous (that's because of properties of stereographic projection, the closer arguments get to infinity in $\mathbb{R}^n$ the closer values get to $(1,0,\ldots,0)$) and "onto". Also
$$f(x)=f(y)\mbox{ if and only if }x=y\mbox{ or }\lVert x\rVert = \lVert y\rVert = 1$$
In particular $f$ induces the mapping
$$\overline{f}:D^n\ /\sim\to S^n$$
$$\overline{f}([x])=f(x)$$
This map is continous, one-to-one and "onto". Since $D^n$ is compact then so is $D^n\ /\sim$ (you only need to verify that it is Hausdorff) and thus $\overline{f}$ is also closed. Therefore the inverse is continous so $\overline{f}$ is a homeomorphism. $\Box$
