Give in the $ \mathbb{R}^n$ euclidean space infinitely many vectors, from which any $n$ is linearly independent, neither of both are orthogonal.
-I still have no idea how to give these vectors, I would appreciate any help. Should I maybe rotate these vectors with "very little angles"?
First a quick and easy proof:
Assume $n\geq2$, and consider the vectors $${\bf x}_k:=(1,k,k^2,k^3,\ldots,k^{n-1})^\top\in{\mathbb R}^n\qquad(k\geq1)\ .$$ If you pick any $n$ different ${\bf x}_k$'s their coordinates form a Vandermonde matrix $V$, hence ${\rm det}(V)\ne0$. This implies that the $n$ chosen vectors are linearly independent.
Now an existence proof not using determinants:
Assume $n\geq2$. Let $A\subset{\mathbb R}^n$ be a finite set of at least $n$ vectors such that any $n$ elements of $A$ are linearly independent. Any $(n-1)$-subset $S\subset A$ then spans a subspace of dimension $n-1$. The union $\Omega$ of these finitely many proper subspaces cannot be all of ${\mathbb R}^n$. We therefore can choose a vector $p\in{\mathbb R}^n$ which is not in the span of any such $S$. Put $A':=A\cup\{p\}$. I claim that $A'$ still has the desired property.
Proof. If $x_1$, $\ldots$, $x_n$ are $n$ different vectors from $A'$, none of them $=p$, then all $x_k$ are elements of $A$, hence linearly independent by assumption on $A$. If, e.g., $x_n=p$, then the $n-1$ vectors $x_1$, $\ldots$, $x_{n-1}$ form a set $S$ as considered above, and $x_n=p\notin{\rm span}(x_1,\ldots,x_{n-1})$ by construction. It follows that the $n$ vectors $x_1$, $\ldots$, $x_n$ are linearly independent.$\quad\square$