Bieberbach theorem say that every discrete cocompact subgroup in $Isom(\mathbb{R}^n )$ contains a translational subgroup of finite index. (the translations forming an abelian normal subgroup of finite index)
1) Why this subgroup is a lattice in $ \mathbb{R}^n$ ?
2) Can you give me a simple proof in dimension 2 ?.
Let's denote the crystallographic (i.e. discrete and cocompact) by $\Gamma$, and write $\operatorname{Isom}(\mathbb{R}^n) = O(n) \ltimes \mathbb{R}^n$. The first Bieberbach theorem, as stated in [1], is:
If $\Gamma \subset \operatorname{Isom}(\mathbb{R}^n)$ is a crystallographic group then the set of translations $\Gamma \cap (\{I_n\} \times \mathbb{R}^n)$ is a torsion free and finitely generated abelian group of rank $n$, and is a maximal abelian and normal subgroup of finite index.
The fundamental theorem of finitely generated abelian groups then implies that $T \cong \mathbb{Z}^n$, which of course means $T$ is a lattice.
For the proof, I'll refer to [1], the second chapter is dedicated entirely to proving the Bieberbach theorems. I do however doubt that the proof given there will simplify a lot in dimension $2$.
Perhaps of some interest to you is the following theorem (thm 2.2 in [1]) by Zassenhaus:
A group $\Gamma$ is isomorphic to a crystallographic group of dimension $n$ if and only if $\Gamma$ has a normal, free abelian subgroup $\mathbb{Z}^n$ of finite index which is a maximal abelian subgroup of $\Gamma$.
[1] Szczepanski, Andrzej. Geometry of crystallographic groups. Vol. 4. World Scientific, 2012.
You can probably find the answers in Chapters 24, 25 and 26 of Armstrong's Groups and symmetry, which gives a pretty elementary treatment of the subject (Chapter 26, in particular, concentrates on dimension $2$).
