To prove there is only a finite number of subgroups in G if $H$ is subgroup of  finite index in $G$. prove that there is only a finite number of distinct subgroups in $G$ of form $aHa^{-1}$.
 A: Every such subgroup $aHa^{-1}=aHHa^{-1}$ is of the form $bH Hc$ and there are only a finite number of options for $bH$ and a finite number of options for $Hc$ (in each case there are $[G: H]$ options). So the number of subgroups is bounded by $[G:H]^2$ (this bound is most likely very weak).
A: Since $H$ is a subgroup of finite index, $G/H$ has a finite numbers of elements.  Let $S_H=\{ aHa^{-1},a\in H\}$, consider the map $f:S_H\rightarrow G/H$ such that $f(a)$ is the class of $a$. We show that $f$ is injective. Suppose that $f(a)=f(b)$, this implies that $b\in aH$, thus $b=ah, h\in H$, $bHb^{-1}=ahH{(ah)}^{-1}=a(hHh^{-1})a^{-1}=aHa^{-1}$, this implies that $aHa^{-1}=bHb^{-1}$. Thus $f$ is injective, since $G/H$ is finite, $S_H$ is finite.
A: Take any coset for $H$ in $G$: e.g., $aH= \{ah \mid h\in H \}$. For any two arbitrary $ah_1$ and $ah_2$ from this coset the conjugate subgroups $ah_1 H(ah_1)^{-1}$ and $ah_2 H(ah_2)^{-1}$ are the same (it is the same as the subgroup $aHa^{-1}$.  SO number of distinct subgroups is limited by the number of cosets $aH$. Your hypothesis says this index of $H$ is $G$ is finite QED.
