$S^n$ is a quotient of closed unit disc in $\mathbb{R}^n$ I am trying to write rigorous proof of the statement in title, since this is really a simple statement but this will check much about my understanding of quotient spaces. 
Let $B$ denote open unit disc in $\mathbb{R}^n$ and $D$=closed unit disc in $\mathbb{R}^n$. Then we have disjoint union:
$$D=B\cup \partial B.$$
(1) Let $B\cup\{\infty\}$ denote the one-point compactification of $B$; it is homeomorphic to one point compactification of $\mathbb{R}^n$ (since $B\cong \mathbb{R}^n$), which is $S^n$.
(2) Define a map $f:(B\cup \partial B)\rightarrow B\cup\{ \infty\}$ (keeping in mind $B\cup\{\infty\}\cong S^n$) by 
$$f(x)=x \mbox{ if }x\in B \mbox{ and } f(\partial B)=\infty.$$
It is easy to prove that $f$ is continuous; obviously it is surjective. 
(3) Next I want to prove that $f$ is open (or closed) map. This ensures that the topology of $B\cup \{\infty\}$ is the quotient topology through $f$. This I was unable to prove. I was trying to prove it for basic open (closed) sets in the closed disk; but still couldn't succeed. Any hint for this?

There could be different proofs posted on this site of statement in the title; but the above three steps describe a way I was trying to prove myself, and where I get stucked. Here in (3) I was lacking in understanding of quotient topology. For proof of (2), one can use structure of open sets in one-point compactification; am I right?
 A: $\mathbb{S}^n \simeq \alpha\mathbb{R}^n$ where $\alpha X$ denotes the Aleksandrov (one-point) compactification of a locally compact space $X$. This is a classical fact proved by the stereographic projection.
A handy fact I will use (proof here):

Theorem If $X$ is locally compact and Hausdorff and $Y$ is compact Hausdorff such that for some $p \in Y$, $X \simeq Y \setminus \{p\}$, then $\alpha X \simeq Y$.

Also if $D^n = \{x \in \mathbb{R}^n : \|x\| \le 1\}$ is the closed unit ball (disk I would reserve for $n=2$, really), then $U=\operatorname{int}(D^n) \simeq \mathbb{R}^n$  and $A = \partial D^n = \mathbb{S}^{n-1}$ is a closed subset, and if $q: D^n \to Y= D^n / A$ is the quotient map that identifies $A$ to a point, then $Y$ is compact and Hausdorff (in the quotient topology) and $q|U$ is a homeomorphism between $U$ and $q[U] \subseteq Y$. So by applying the above theorem to the identified point $q[A]$ of $Y$ we see that $$Y \setminus \{q[A]\} \simeq U \simeq \mathbb{R}^n$$ so
$$Y \simeq \alpha \mathbb{R}^n \simeq \mathbb{S}^n$$ and so the $n$-sphere is a quotient of $D^n$, as claimed.
