A finite group is abelian iff all its irreducible characters have dimension one (hence are linear). A common proof uses that the number of irreducible representations equals the number of conjugacy classes (hence $|G|$ in an abelian group) and that $\sum_{i=1}^k \chi_i(1)^2 = |G|$, which for $k = |G|$ is equivalent with $\chi_i(i) = 1$ (as $\chi_i(1) \ge 1$) for all $i = 1,\ldots, k$.
Is there a more computational proof? Going along the lines, let $\pi : G \to V$ be a representation of an abelian finite group with $\mbox{dim}(V) > 1$ and suppose $1 < W < V$, then $\pi(g)W \le W$...
Is there any way to continue this meaningful? I want to know if the fact that elements of $G$ commute could be used directly in a proof of the above stated property, without using such "higher level" theorems?
$G$ abelian implies irreducibles have dimension 1: Let $V$ be an irreducible representation of $G$ (necessarily finite-dimensional). By Schur's lemma, every element $g \in G$ acts by multiplication by a scalar, so every subspace of $V$ is $G$-invariant. Hence we must have $\dim V = 1$.
Irreducibles have dimension $1$ implies $G$ abelian: This is the hard direction, and one way or another you need a result that tells you that $G$ has "enough irreducibles." I'll use the fact that a finite group $G$ acts jointly faithfully on its irreducible representations. Since these all have dimension $1$, every $g \in G$ acts by a scalar on all of its irreducibles, hence every $g, h \in G$ commute in all of the irreducibles. Hence they commute.