Could someone help me with an simple example of a profinite group that is not the p-adics integers or a finite group? It's my first course on groups and the examples that I've found of profinite groups are very complex and to understand them requires advanced theory on groups, rings, field and Galois Theory. Know a simple example?
Last, how to prove that that $\mathbb{Z}$ not is a profinite group?
Yes. Let $G = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ be the direct product of (countably) infinitely many copies of the cyclic group of order $2$. This profinite group, sometimes called (well, by me at least) the Bernoulli group, occurs naturally in probability theory. As a topological space it is homeomorphic to the standard Cantor set.
(For that matter, any Cartesian product of finite groups is a profinite group, and this is an important example, because any profinite group is a closed subgroup of such a product.)
As for your second question, a profinite group is in particular a compact (Hausdorff) topological group and thus carries a Haar measure, i.e., a translation-invariant probability measure. Thus it cannot be countably infinite, and in particular $\mathbb{Z}$ is not (or more precisely, cannot be endowed with the structure of) a profinite group.
One way to get a profinite group is to start with any torsion abelian group $A$, and take $\hom(A, \mathbb{Q}/\mathbb{Z})$. This acquires a topology as the inverse limit topology: it is the inverse limit of $\hom(A_0, \mathbb{Q}/\mathbb{Z})$ for $A_0 \subset A$ a finitely generated (and necessarily torsion) subgroup.
In fact, this gives an anti-equivalence between profinite abelian groups and torsion abelian groups, which is sometimes useful (it means, for instance, that a filtered inverse limit of profinite abelian groups is always exact, since the corresponding fact for filtered direct limits is true in abelian groups).
