Homeomorphism between open unit ball and $\mathbb R^n$

Injectivity

$F(x)=F(y)$, with $x\ne0$ and $y\ne0$ means

$$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}= \tan\biggl(\frac{\pi \|y\|}{2}\biggr)\frac{y}{\|y\|} $$

So we can assume $x\ne0$ and $y\ne0$. Then, taking norms, $$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)=\tan\biggl(\frac{\pi \|y\|}{2}\biggr) $$ because $x/\|x\|$ has norm $1$ and the same for $y/\|y\|$; moreover $\tan t>0$ if $t>0$. By the injectivity of the tangent you get $$ \frac{\pi \|x\|}{2}=\frac{\pi \|y\|}{2} $$ so $\|x\|=\|y\|$ and, finally $x=y$.

It's clear that $F(x)=0$ if and only if $x=0$, so the proof is complete.

Surjectivity

$0=F(0)$, so we only need to show that, for $z\ne0$, we can find $x$ with $F(x)=z$. We should have $$ z=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|} $$ so $$ \|z\|=\tan\biggl(\frac{\pi \|x\|}{2}\biggr) $$ hence $$ \frac{\pi\|x\|}{2}=\arctan\|z\| $$ that is, $$ \|x\|=\frac{2\arctan\|z\|}{\pi}. $$ Thus the candidate is $$ x=\frac{2\arctan\|z\|}{\pi\|z\|}z $$ Verify it's the right one.

Continuity

The continuity of $F$ is obvious in the points different from $0$, because it's obtained by continuous functions. Since your domain consists of vectors with norm $<1$, there's no problem with the tangent function, because you consider only arguments in $(0,\pi/2)$. The continuity of the inverse is obvious as well outside $0$

Are $F$ and its inverse continuous at $0$?

A well known fact is that $\lim_{x\to0}F(x)=0$ if and only if $\lim_{x\to0}\|F(x)\|=0$; now $$ \|F(x)\|=\biggl\|\tan\biggl(\frac{\pi\|x\|}{2}\biggr)\frac{x}{\|x\|}\biggr\| =\tan\biggl(\frac{\pi\|x\|}{2}\biggr) $$ that obviously satisfies the requested property. The same for the continuity at $0$ of $F^{-1}$.