Let us prove this in a series of steps. Firstly, we obviously need to understand the representations of abelian groups. The following exercise is important for this understanding:
Exercise 1 Prove that if $G$ is an abelian group, then every irreducible representation of $G$ is $1$-dimensional. (Hint: use Schur's lemma on the $G$-intertwining maps from a (complex) irreducible representation of $G$ to itself.)
Let us now consider a group $G$ with an abelian subgroup $H$ of index $2$. If $(\pi,V)$ is an irreducible representation of $G$, then $\pi$ induces a representation of $H$ by restriction which we denote by $(\pi_H,V)$. We consider two cases:
(1) If $(\pi_H,V)$ is irreducible, then $V$ is $1$-dimensional and the proof is complete.
(2) If $(\pi_H,V)$ is not irreducible, then there exists a minimal $H$-invariant subspace $W\subseteq V$. The following exercise allows us to understand $V$ in terms of $W$:
Exercise 2 If $g\in G$ and $g\not\in H$, then prove that $V=W+gW$. (Hint: remember that $(\pi,V)$ is an irreducible representation of $G$!)
Finally, we can determine a bound on the dimension of $V$:
Exercise 3 Prove that the dimension of $V$ is at most $2$. (Hint: the minimality of $W\subseteq V$ as a $H$-invariant subspace implies that $(\pi_H,W)$ is an irreducible representation of $H$; use Exercise 1.)
You could argue that the proof above is somewhat complicated. In fact, one can rewrite the above proof in a more conceptual manner which allows one to see the real idea of the proof. The point is essentially the relationship between the representations of a group and the induced representations of subgroups and, where applicable, quotient groups.
I hope this helps!