The theorem about the equivalence of the norms 
The theorem about the equivalence of the norms says that every two
  norms in $ \Bbb R^n$ are equivalent. It can be written as $c_1||x||
 \le ||x||_2 \le c_2 ||x||$. I struggle with imaging it and don't know
  how to prove it (or I don't understand that proof).



*

*Does it mean that it doesn't matter which norm I would use for my given vector space if I want to measure something there or that If I am using one type of the norm it is easy to tranfer to another? The Lipschitz continuity says if I have the linear mapping between two metric spaces such that $f: (M, d_m) \to (N, d_n)$ than we have some constant $K \gt 0$ for every $x,y \in M$ such that $d_N(f(x), f(y)) \le Kd_M(x-y)$. So transfering between the norms then means the linear transformation (homomorphism). 

*I have alson read that $f^{-1}$ exists and that the mapping works for the finite (bounded) vector spaces, the first condition I understand, the second I don't. 

*Why I should compare all the norms with euclidean norms?
 A: The result you're quoting holds true for finite-dimensional vector spaces (there are some norm equivalences on infinite-dimensional spaces but you usually have to prove/verify them first).  You're right in your first question: if two norms are equivalent then you can swap them in your working so long as you remember the constant multiplier needed.  What this allows you to do is to choose the norm best suited to the analysis you're doing at any given time.
For example, you might have started with the Euclidean norm on the unit disc and then realised that you're in a space that looks pretty hyperbolic, so you might find that switching to the norm generated by the Poincaré metric ($d_P(z_1, z_2) = \tanh^{-1}\left| \frac{z_1-z_2}{1-z_1\bar{z_2}}\right|$) simplifies many of your calculations.
For the second question, norm-equivalence holds for finite-dimensional vector spaces essentially because the unit ball in those spaces is compact.  There's an excellent discussion of it here (Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces) but (very loosely speaking) in finite-dimensional spaces there's not enough room for the norms to misbehave.
And finally: you don't have to compare all norms with the euclidean ones.  Choose your favourite norm and compare other norms with that.  However, you'll probably find that most of the literature uses Euclidean norms because they're well understood and easily represented in the standard basis on a finite-dimensional vector space.
