If $H,K < G$ and $(G:H), (G:K)$ are finite, prove $(G: H \cap K)$ is finite. If $H,K < G$ and $(G:H), (G:K)$ are finite, prove $(G: H \cap K)$ is finite.
I wrote a proof but I'm not sure whether it works. Please, tell me if it is correct. Here it goes:
If $(G:H)$ and $(G:K)$ are finite, then we can find $n$ distinct $g$'s in $G$ such that $g_{h_i}H,\,i = 1,2, \ldots, n$ produce all the left cosets of $H$ in $G$ and $m$ distinct $g$'s such that $g_{k_j}K,\,j = 1,2, \ldots, m$ produce all the the left cosets of $K$ in $G$. We then form sets based on these elements. What I mean is, we form $\overline{g_{h_1}}, \overline{g_{h_2}}, \ldots, \overline{g_{h_n}}$, so that if $g \in \overline{g_{h_1}}$, for example, these $g$'s are going to produce the same left coset. We do the same for $K$. Consider now the intersection, $H \cap K$. For any $g \in G$, this element belongs to only one of $\overline{g_{h_1}}, \ldots, \overline{g_{h_n}}$ and to only one of $\overline{g_{k_1}}, \ldots, \overline{g_{k_m}}$. We then have at most $mn$ distinct pairs, $\overline{g_{h_j}}$ and $\overline{g_{k_i}}$. In other words, $H \cap K$ has at most $mn$ distinct left cosets, which shows that $(G: H\cap K) < \infty$.
It may be a little confusing. I will clarify the steps more if needed.
$\textbf{Edit to answer Brian Moehring's question}$: I thought a little bit more about the last part of the proof. For example, $g \in \overline{g_{h_1}}$ is going to generate elements of a left coset of $H$, say  $H_1$. But, because we are considering  $H \cap K$, it's going to be contained in $H_1$. Therefore, we need to consider in which of the $\overline{g_{k_1}}, \ldots, \overline{g_{k_m}}$ it belongs to. Let's say it belongs to $\overline{g_{k_1}}$. Then is going to generate elements of the left coset $K_1$. The pair $(\overline{g_{h_1}}, \overline{g_{k_1}})$ then forms the left coset $H_1 \cap K_1$.
$\textbf{Edit (I wrote another proof based on the comments and suggestions made here)}$: If $(G:H)$ and $(G:K)$ are finite, then we can find $n$ distinct $g$'s in $G$ such that $g_{h_i}H,\,i = 1,2, \ldots, n$ produce all the left cosets of $H$ in $G$ and $m$ distinct $g$'s such that $g_{k_j}K,\,j = 1,2, \ldots, m$ produce all the the left cosets of $K$ in $G$. Now, consider the intersection $H \cap K$. For an arbitrary $g \in G$, we have $g(H \cap K) = gH \cap gK = g_{h_i}H \cap g_{k_j}K$ for some $i,j$. From this we can conclude that $(G: H \cap K) < \infty$, because there are only, at most, $mn$ distinct sets of the form $g_{h_j}H \cap g_{k_j}K$.
 A: This argument is, in spirit, closely related to the Third (or Second, depending how you number them) Isomorphism Theorem.
Theorem. $(H:H\cap K)\leq (G:K)$, in the sense of cardinality, with equality if $G=HK$. In the case where $(G:K)$ is finite, equality holds if and only if $HK=G$.
Proof. I claim that $h_1(H\cap K) = h_2(H\cap K)$ if and only if $h_1K = h_2K$. Indeed:
$$\begin{align*}
h_1K = h_2K &\iff h_2^{-1}h_1\in K\\
&\iff h_2^{-1}h_1\in H\cap K\\
&\iff h_1(H\cap K) = h_2(H\cap K).
\end{align*}$$
Thus, the map from $\{h(H\cap K)\mid h\in H\}$ to $\{gK\mid g\in G\}$ given by sending $h(H\cap K)$ to $hK$ is both well-defined and one-to-one. Thus, $(H:H\cap K)\leq (G:K)$ (in the sense of cardinality).
If $HK=G$, then every element of $\{gK\mid g\in G\}$ can be written as $hkK=hK$ for some $h\in H$, and thus the map above is a bijection, yielding equality. If $(G:K)$ is finite and we have equality, the injection above must be a bijection, and so every coset $gK$ is of the form $hK$ for some $h\in H$. Thus, given $g\in G$, there exists $h\in H$ and $k\in K$ such that $g=hk\in hK$, hence $G\subseteq HK$, which implies the equality. $\Box$
Corollary. $(G:H\cap K)\leq (G:H)(G:K)$ in the sense of cardinality.
Proof. $(G:H\cap K)= (G:H)(H:H\cap K) \leq (G:H)(G:K)$, in the sense or cardinality. $\Box$
Corollary. If $H$ and $K$ have finite index in $G$, then $H\cap K$ has finite index in $G$, and moreover $(G:H\cap K)\leq (G:H)(G:K)$, with equality if and only if $G=HK$.
Proof. The inequality was already proven. We get equality if and only if $(H:H\cap K)=(G:K)$, which holds in this situation if and only if $G=HK$. $\Box$
Why do I say it is related in spirit to the Third Isomorphism Theorem? That's the one that says that if $K$ is normal, then $HK/K$ is isomorphic to $H/H\cap K$. The bijection between these two is precisely the one that sends $h(H\cap K)$ to $hK$. If we knew $HK$ is a subgroup, we could show that $(HK:K)=(H:H\cap K)$, and then use that $HK\leq G$ to get $(H:H\cap K)\leq (G:K)$.
