To use 2.3, suppose $G$ is simple with irreducible character $\chi$ of degree $2$. If $\rho\colon G\to GL_2(\mathbb{C})$ affords $\chi$, then $\ker\rho\unlhd G$ must be trivial, since $\rho$ is not the trivial rep, so $\rho$ is an injection.
Now $G$ can't be abelian, so $[G,G]\neq 1$, so by simplicity, $G=[G,G]$. The linear characters of $G$ are in bijection with $G/[G,G]$, which has order $1$, so the only linear character is the trivial character. But by 2.3, $\det\chi$ is a linear character, hence it must be trivial. So $\det\rho(g)=1$ for all $g\in G$.
Now the degree of an irreducible character divides $|G|$, so $2$ divides $|G|$. By Cauchy, there exists $g\in G$ of order $2$. Then $\rho(g)$ has order $2$, and is similar to a $2\times 2$ diagonal matrix whose diagonal entries are 2nd roots of unity, i.e., $\pm 1$. So necessarily $\rho(g)$ is similar to $-I_2$, since it must have determinant $1$, hence actually $\rho(g)=-I_2$. Then $\rho(g)\rho(x)=\rho(x)\rho(g)$ for all $x\in G$, so since $\rho$ is injective, $g$ is central in $G$. So finally $\langle g\rangle\unlhd G$, and is proper, since $|G|\neq 2$, else $G$ would be abelian. But this contradicts simplicity.
Let $A \unlhd G$ and $\chi$ a monomial irreducible (complex) character say $\chi=\lambda^G$, with $\lambda$ linear character of a subgroup $K$ of $G$. Consider the subgroup $KA$. Observe that $\lambda^{KA}$ is irreducible. By using Problem (5.2) in Isaacs' book (all references are in Character Theory of Finite Groups), we have $(\lambda^{KA})_A=(\lambda_{K \cap A})^A$. Since $A$ is abelian, we can find a linear character $\mu$ of $A$, such that $\mu_{K \cap A}=\lambda_{K \cap A}$. Now apply (6.17) Corollary (Gallagher): we must have
$$(\lambda^{KA})_A=(\lambda_{K \cap A})^A=\sum_{\nu \in Irr(A/K \cap A)} \mu \nu$$
Observe that all $\nu$ and hence $\mu\nu$ are linear, since $A$ is abelian. The corollary also says that all the $\mu\nu$ are distinct and are all of the irreducible constituents of $(\lambda_{K \cap A})^A$. Hence $(\lambda^{KA})_A$ is the sum of distinct conjugates of the $\mu$. Now apply Clifford's Theorem ((6.11) Theorem), there must be a linear character $\psi \in I_{KA}(\mu)$, the inertia group of $\mu$, with $[\psi_A,\mu] \neq 0$, such that $\psi^{KA}=\lambda^{KA}$. Hence $\chi=\lambda^G=(\lambda^{KA})^G=(\psi^{KA})^G=\psi^G$ by transitivity of induction (see Problem (5.1)). So $H=I_{KA}(\mu)$ is the requested subgroup.
Best Answer
If $A$ is a normal abelian $p$-complement of $N$, then gcd$(|A|,|N:A|)=1$, implying that $A$ is characteristic in $N$. So $A$ char $N \lhd G$, hence $A \lhd G$. And $|G:A|$ is a $p$-power, $p \nmid |A|$, so at this point you could apply the theorem of Schur-Zassenhaus, providing a complement in $G$ to $A$. But you do not have to invoke that heavy theorem: $G/A$ is a $p$-group, so if $S \in Syl_p(G)$, then $SA/A=G/A$. Hence $G=SA$ and of course $S \cap A=1$.