Let $G$ be a finitely generated group, and $H \subset G$ a subgroup of finite index. Show that $H$ is finitely generated.

Question

Problem:

Let $G$ be a finitely generated group, and $H \subset G$ a subgroup of finite index. Show that $H$ is finitely generated.

My work:

I wrote this proof:

Consider any finitely generated group $G = \langle x_1, x_2, \ldots, x_n \rangle$ and a subgroup $H$ with a finite index. For some $g_i \in G - H$, $G/H = \{H, g_1 H, \ldots, g_m H\}$ so consider the set $S = \{1, g_1, \ldots, g_m\}$, since $G/H$ is a partition of $G$, for all $x \in G$ there is some $h \in H$ and some $s \in S$ such that $x = sh$. Since $G$ is finitely generated any element of $H$ can be expressed as $x_{i_1}^{\alpha_1} \ldots x_{i_k}^{\alpha_k} = (s_{i_1}h_{i_1})^{\alpha_1} \ldots (s_{i_k}h_{i_k})^{\alpha_k}$

But, as you can see, it is incomplete since I was wondering if those were the generators of $H$ so I came up with a concrete example to see what happened.

I thought of the group $D_4$, consider the $90^{\circ}$ rotation $r$ and any reflection $h$, $D_4 = \langle r, h \rangle$ and the subgroup $G = \{1, h, r^2 h, r^2\} \subset D_4$. We can check that $D_4/G = \{G, rG\}$ so consider the set $A = \{1, r\}$. Since $D_4/G$ is a partition of $D_4$ we know that for all elements $d \in D_4$, for some $a \in A$ and some $g \in G$, $d = ag$ and thus $D_4 = \langle r, h \rangle = \langle a_1 g_1, a_2 g_2 \rangle = \langle r 1, 1h \rangle$. After finishing this example I can't see why it is useful knowing all of the above because I can't see a way to conclude that $G = \langle r^2h, h\rangle$ is indeed finitely generated.

In my proof I took that approach because I was given the following hint:

Choose a finite subset $S \subset G$ containing one representative of each coset of $H$, so every element of $G$ is the product of an element of $S$ and an element of $H$. Given a word in the generators of $G$ and their inverses, how do you rewrite it as the product of an element of $S$ and an element of $H$?

I would appreaciate any hints on how to continue my proof.

I already saw subgroups of finitely generated groups with a finite index but I don't understand how to conclude that $h_{i_1} \ldots h_{i_k}$ are the generators of $H$.

Thanks in advance.