Every open ball of a normed vector space $E$ its homeomorph to the entire space $E$. I have some questions about this proof that "Every open ball of a normed vector space $E$ its homeomorph to the entire space $E$.":
By the example (12), we just have to consider the ball $B(0,1)$, we just have to show an homeomorphism $E\to B$. Choose $f(x) = \frac{x}{1+|x|}$. Since $|f(x)| = \frac{|x|}{1+|x|}<1 \ \ \forall x\in E$, we see that $f$ is continuos from $E$ to $B$. WHY???
Define now $g:B\to E$, where $g(y) = \frac{y}{1-|y|}$. Since $|y|<1$ for all $y\in B$, we see that $g$ is continuous. (WHY?)
Also, we can see that $f(g(y)) = y$ and $g(f(x)) = x$ for all $y\in B$ and $x\in E$, then $g = f^{-1}$ and $f$ is an homeomorphism.
example 12 says that two open balls in a normed vector space are homeomorphic. How does it help in the proof above?
 A: I think that when he says 

$f$ is continuous from $E$ to $B$.

he intends to mean that the important part which was justified was:

(..) to $B$.

and assumes continuity is a trivial matter (which is, since it is division by $\frac{1}{1+\Vert x \Vert }$. Since the norm is continuous, multiplication by scalar is continuous, inversion is continuous in $\mathbb{R}$ and sum is continuous in $\mathbb{R}$, we have that the function is continuous).
Likewise, when he says that $g$ is continuous, I think he just intends to mean that it is well-defined.
Now, as to why two open balls are homeomorphic helps in this case, it is just a matter of working in the particular case of the open ball centered in $0$ with radius $1$, and then composing with a homeomorphism to have a homeomorphism from the whole space to any ball.

Just for the sake of completeness, opening up why $\frac{1}{1+\Vert x \Vert}x$ is continuous:
Consider the functions $\Vert \cdot \Vert: E \rightarrow \mathbb{R}$, $x \mapsto \Vert x \Vert$; 
$m:\mathbb{R} \times V \rightarrow V,$ $(\lambda, x) \mapsto \lambda x$;
$i: \mathbb{R}\backslash\{0\} \rightarrow \mathbb{R},$ $r \mapsto \frac{1}{r}$;
$t_1: \mathbb{R} \to \mathbb{R}$, $r \mapsto 1+r$,
we have that $f=m \circ \bigg( \big(i \circ t_1 \circ \Vert \cdot \Vert \big) \times Id\bigg)$, which is a composition of continuous functions.
