If a group $G$ contains a subgroup ($\neq G$) of finite index, it contains a normal subgroup ($\neq G$) of finite index.
I have written a solution below. I used proposition 4.8 section 2.4 and portion of proof of corollary 4.10 section 2.4 to prove above exercise. Statement of proposition 4.8 is included in answer. My proof don’t reprove any results which have been proven in that section.
Proposition 4.8. Let $H$ be a subgroup of a group $G$ and let $G$ act on the set $S$ of all left cosets of $H$ in $G$ by left translation. Then the kernel of the induced homomorphism $\varphi:G\to A(S)$ is contained in $H$, where $A(S)$ is a group of all permutation of $S=\{gH\mid g\in G\}$.
Proof: Suppose $H(\neq G)$ be a subgroup of $G$ with $[G:H]=n\in \Bbb{N}$. Let $K=\ker \varphi$. By proposition 4.8, $K\subseteq H$. Since $[G:H]=n$, it is easy to check $A(S)\cong S_n$. By first isomorphism theorem, $G/K\cong \text{Im} \varphi \leq A(S)\cong S_n$. So $|G/K|=|\text{subgroup of }A(S)|$ divides $|S_n|=n!$. Which implies $|G/K|=[G:K]$ is finite, $1\leq [G:K]\leq n!$ to be precise. Clearly $K\neq G$, since $K\subseteq H\subset G$. Thus $\exists K(\neq G)$ subgroup of $G$ such that $K\subseteq H$, $K\lhd G$ and $[G;K]$ is finite.