Let $K<H<G$ be a chain of subgroups. Let $R \subset G$ be a system of representatives for $G/H$ and let $S \subset H$ be a system of representatives for $H/K$.
I have to show the following:
(a) Show that the map $R \times S \rightarrow RS$ given by $(r,s) \mapsto rs$ is a bijection.
(b) Show that $RS = \{rs | r \in R, s \in S\}$ is a system of representatives for $G/K$ and conclude that $[G:K] = [G:H][H:K]$.
I am completely lost on how to prove both (a) and (b) and would appreciate any help.
May as well write an answer since part (a) has been done in the comments and (b) is a natural consequence of (a).
Let's give the map in part (a) a name: $\phi : R \times S \to RS$, where $\phi(r,s) = rs$.
We first show that $\phi$ is an injection. Suppose $r_1 s_1 = r_2 s_2$. Then $r_2^{-1} r_1 = s_2 s_1^{-1}$. Since $s_1$ and $s_2$ are in $S$, which is a subset of $H$, and $H$ is a subgroup, it follows that $s_2 s_1^{-1} \in H$, so $r_2^{-1} r_1 \in H$. This means that $r_1 H = r_2H$, in other words $r_1$ and $r_2$ are in the same coset of $H$. Since $R$ contains exactly one representative from each coset of $H$, we must have $r_1 = r_2$. But then $r_2^{-1} r_1 = 1$, the identity element, and this means that $s_2 s_1^{-1}$ is also the identity, so $s_1 = s_2$.
Now we show that $\phi$ is a surjection. This is immediate from the definition, because an arbitrary element of $RS$ is of the form $rs$, where $r \in R$ and $s \in S$, and therefore $rs$ is the image under $\phi$ of the element $(r,s) \in R \times S$.
We conclude that $\phi$ is a bijection.
For part (b), define $\psi : R \times S \to G/K$ by $\psi(r,s) = rsK$.
We show that $\psi$ is an injection. Suppose that $r_1 s_1 K = r_2 s_2 K$. Since $s_1 K \subset H$ and $s_2 K \subset H$, it follows that $r_1 s_1 K \subset r_1 H$ and $r_2 s_2 K \subset r_2 H$. Therefore, $r_1 H$ and $r_2 H$ are not disjoint, so they must be equal: $r_1 H = r_2 H$. This implies that $r_1 = r_2$ by definition of $R$. Let us set $r = r_1 = r_2$. Then we have $rs_1 K = rs_2 K$, so $rs_1 \in rs_2 K$, and therefore $s_1 \in s_2 K$, which means that $s_1 K = s_2 K$. By definition of $S$, this means that $s_1 = s_2$.
We show that $\psi$ is a surjection. If $g$ is an arbitrary element of $G$, then it lies in some element of $G/H$, say $g \in rH$. Then $r^{-1}g \in H$, so $r^{-1}g$ lies in some element of $H/K$, say $r^{-1}g \in sK$. Then $g \in rsK$, so $gK = rsK = \psi(r,s)$.
Finally, note that the statement "$RS$ is a system of representatives of $G/K$" is equivalent to the statement "there is a bijection from $RS$ to $G/K$ which maps $rs \to rsK$". But this is now clear: $\psi \circ \phi^{-1} : RS \to G/K$ is the desired bijection.
We want to conclude that $[G:K] = [G:H][H:K]$.
Note that this follows immediately from what we have already done.
Let's assume that $[G:K]$ is finite. Then, by definition, $[G:K]$ is the number of elements in $G/K$. We have shown that there is a bijection between $G/K$ and $RS$, so $[G:K]$ is also equal to the number of elements in $RS$, which we may denote as $|RS|$. Similarly, $[G:H] = |G/H| = |R|$ and $[H:K] = |H/K| = |S|$. So the equation $[G:K] = [G:H][H:K]$ is the same as $|RS| = |R| |S|$. This in turn is true because we have a bijection between $RS$ and $R \times S$, and the size of $RS$ is $|RS|$, and the size of $R \times S$ is $|R| |S|$.
If $[G:K]$ is infinite, then one or both of $[G:H]$ or $[H:K]$ must also be infinite, again because of the bijection between $RS$ and $R \times S$. So the equation holds (trivially) in that case as well: both sides are $\infty$.
Bungo handed me a) on a platter. But here are my thoughts for b):
Since we don't know if any of the groups are finite, let $R=\{g_1,g_2,...\}$ and $S=\{h_1,h_2,...\}$ where $$\bigcup_{r\in R}rH=G,\quad\text{and}\quad\bigcup_{s\in S}sK=H$$ by definition of $R,S$ being sets of representatives. Now consider any $g\in G$. Then $g=r_{\alpha}H$ for some $\alpha\in\mathbb{Z}$, hence there is some $h\in H$ s.t. $g=r_\alpha h$. Now consider $h\in H$. Then $h=s_\beta K$, some $\beta\in\mathbb{Z}$, so $\exists k\in K$ s.t. $h=s_\beta k$. Then we can simply compute $$g=r_\alpha h=r_\alpha s_\beta k$$ Hence $g\in r_\alpha s_\beta K$ where $r_\alpha s_\beta =g'$ for some $g'\in G$. Since $g$ was arbitrary we can conclude that every $RS$ contains a representative for every $g\in G$.
