The unit ball $S^n$ is homeomorphic to the one-point compactification of $R^n$ I saw many different proofs in the community of the statement in the title. And I want to demonstrate my proof of the general case. 
I have tried to show the one-point compactification is homeomorphic to $S^n$ directly, but encountered difficulty in showing the continuity. So I found another clean proof which I want to share with you.
 A: Let $p={(0,0,...,0,1)} \in R^{n+1}$, and consider the stereographic projection $$ 
f: S^n\setminus\left\{ p\right\} \rightarrow R^n
$$
defined by $$
f(x_1,x_2,...x_{n+1})=\frac{1}{1-x_{n+1}}(x_1,x_2,...x_n)
$$
Now we can directly show that $f$ is a bijective map by direct calculation of the coordinates. 
Let $\pi_i:R^n\rightarrow R$ be the projection map. It suffices to show that $\pi_i \circ f$ is continuous for the continuity of $f$. So let $(a,b) $ be an open interval in $R$, we have
$f^{-1}(a,b)=\left\{ x_i\in R: a<\frac{x_i}{1-x_{n+1}}<b 
\right\}\cap S^n$,
which is open in $S^n$. So $f$ is continuous.
Consider the one-point compactification $S^n$ of $S^n\setminus\left\{p \right\}$. (One-point compactification is unique up to homeomorphism). We can apply Tietze Extension theorem to get $f_0$, a continuous extension of $f$ on $S^n$. Since $S^n\setminus\left\{p \right\}$ is dense in $S^n$, we have the continuous extension being unique. 
Claim: $f_0$ is bijective.
To prove the claim, it suffices to check whether $f_0(p)=a$ for some $a \in R^n$. Suppose so, we consider the sequence $\zeta_n=(\frac{1}{n},0,0,...,0,\sqrt{1-\frac{1}{n^2}})\in S^n\setminus\left\{p \right\}$.
$f_0(\zeta_n)=f(\zeta_n)=(\frac{1}{n-\sqrt{n^2-1}},0,0,...,0)$.
But $$\lim_{n\to\infty} \frac{1}{n-\sqrt{n^2-1}}=\lim_{n\to\infty}(n+\sqrt{n^2-1})$$ does not exist in $R^n$.
So we must have $f_0(p)=\infty$.
Now $f_0$ is a continuous bijection from a compact space $S^n$ to a Hausdorff space $R^n \cup \left\{\infty\right\}$. So it is a homeomorphism.
A: Here's a more general idea without Tietze Extension theorem: Assume that $Y$ is a compact and Hausdorff space and let $y$ be a point in $Y$. Suppose $Y\backslash \{y\}$ is not compact, we prove that $Y$ is homeomorphic to $(Y\backslash \{y\})^+$. If this is true, we can use the fact that $S^n\backslash \{y\}$ is homeomorphic to $R^n$ to derive our conclusion.
For the above lemma, since $Y\backslash \{y\}$ is not compact, then $(Y\backslash \{y\})^+$ is compact and Hausdorff. We consider the natural bijection $f$ from $(Y\backslash \{y\})^+$ to $Y$. Since $Y$ is compact, we can easily verify that the pre-image of any open set in $Y$ under $f$ is open, therefore $f$ is continuous, and we have proved that $f$ is a homeomorphism since $Y$ is Hausdorff.
