I had to prove that $f: \mathbb{R}^n \ni x\to\left\lVert x \right\rVert$ $\in \mathbb{R}$ is continuous regarding maximum norm.
Is it correct to do the following, i.e. using triangle inequality?
$\left\lVert x \right\rVert$ = $\left\lVert x - y + y\right\rVert$ $\leq $ $\left\lVert x - y \right\rVert$ $+ \left\lVert y \right\rVert$
$\left\lVert y \right\rVert$ = $\left\lVert y - x + x\right\rVert$ $\leq $ $\left\lVert y - x \right\rVert$ $+ \left\lVert x \right\rVert$
After rearranging I get
$|f(x) - f(y) |$ $\leq$ $\left\lVert x-y \right\rVert$
which is why f is Lipschitz continuous, and therefore continuous. (I think?)
Now there is $S := {x \in \mathbb{R}^n}$ : $\left\lVert x \right\rVert_\infty = 1$ which is the unit sphere regarding maximum norm. How can I now conclude from above, that $f$ has a minimum $A := f(x) > 0$ in a point $x$ on $S$?
And is it also possible to conclude out of that, that therefore $\left\lVert x\right\rVert$ $\geq$ $A * \left\lVert x \right\rVert_\infty$ for all $x \in \mathbb{R}^n$?
You did not prove continuity of $f$ w.r.t. $\|x\|_{\infty}$ norm. Write any vector $x$ as $\sum x_ie_i$ where $e_1,e_2,..,e_n$ is the standard basis. Then $f(x) \leq \sum |x_i| |f(e_i)|\leq \|x\|_{\infty} \sum |f(e_i)|$ which proves that $f$ is continuous for the $\|x\|_{\infty}$ norm. Since $S$ is compact in $\|x\|_{\infty}$ norm it follows that $f$ has a minimum value $c$ on $S$. [Observe that $f$ does not vanish at any point of $S$]. Now you can verify that $f(x) \geq c \|x\|_{\infty}$ for all $x$ by considering $\frac x {\|x\|_{\infty}}$.