I got it.
Let $A$ be the set of cosets of $H\cap K$
Let $B$ be the set of cosets of $H$
Let $C$ be the set of cosets of $K$
Define $\phi:A\rightarrow B\times C : g(H\cap K)\mapsto (gH,gK)$.
Note that $\phi$ is well defined and also it's trivial that it is injective
Thus, $|A|\leq |B||C|$.
However I'm really wondering whether my argument is correct or not, since above in the link, people are using group actions and isomorphism theorems and stuffs to prove this..
Let $G$ be the disjoint union of the cosets $Ha_1,...,Ha_n$, and furthermore, let $H$ be the disjoint union of the cosets $Kb_1,...,Kb_m$. For brevity, let $K_{i,j} := Kb_ia_j$ and
$\Phi = \{K_{i,j}: 1 \leqslant i \leqslant m,\,\, 1 \leqslant j \leqslant n\}$
First show that the collection $\Phi$ covers $G$. Supposing $x \in G$, there is some $j$ such that $x \in Ha_j$. In particular, $x = ha_j$ for some $h \in H$. Moreover, because $h \in H$ we also know there is some $i$ such that $h \in Kb_i$, which is to say $h = kb_i$ for some $k \in K$. But then $x = ha_j = kb_ia_j \in K(b_ia_j) = K_{i,j}$. This proves that $\Phi$ covers $G$.
Next, we will show $| \Phi | = mn$. To that end suppose $K_{p,q}=K_{r,s}$. We will show that $p=r$ and $q=s$. Now, because $Kb_p,Kb_r \subset H$, we know that $K_{p,q} \subset Ha_q$ and $K_{r,s} \subset Ha_s$. But by construction $Ha_q,Ha_s$ are disjoint when $q \neq s$. So it must be that $q = s$ and $a_q = a_s := a$. Consequently, we have
$Kb_p = (Kb_p)(a_qa^{-1}) = (K_{p,q})a^{-1} = (K_{r,s})a^{-1} = (Kb_r)(a_sa^{-1}) = Kb_r$.
So again by construction, it must be that $p = r$. This concludes the proof that $| \Phi | = mn$
Finish up with
$[G:K] = |\Phi| = mn = [H:K][G:H] = [G:H][H:K]$.
Also, you have not mentioned anything regarding normality of $H,K$. If you knew $H,K$ were normal subgroups of $G$, then as a @FofX noted, the third isomorphism theorem is applicable:
$(G/K) / (H/K) \cong G/H$.
Thus
$[G:K] $
$=|G/K| $
$= [G/K:H/K] \cdot |H/K| $
$= |(G/K)/(H/K)| \cdot |H/K| $
$= | G/H| \cdot |H/K| $
$= [G:H] \cdot [H:K]$
Where the fourth inequality follows from $|(G/K)/(H/K)| = |G/H|$ (which in turn is a direct result of the third isomorphism theorem.)
Best Answer
We'll show that there exists a bijection between $(H\backslash G)\times(K\backslash H)$ and $K\backslash G$. Let $(a_i)_{i\in I}$ be a collection of representatives for the elements of $H\backslash G$ and $(b_j)_{j\in J}$ a collection of representatives for the elements of $K\backslash H$. Define $\phi:(H\backslash G)\times(K\backslash H)\to K\backslash G$ by $$ \phi(Ha_i,Kb_j)=Kb_ja_i. $$ Let $Kc\in K\backslash G$. Then $Kc\subset Ha_i$ for some unique $a_i$ since $K\leq H$, so $c=ha_i$ for some (also clearly unique given an $a_i$) $h\in H$. We also have that $Kh=Kb_j$ for some unique $b_j$, so $Kc=Kb_ja_i=\phi(Ha_i,Kb_j)$, and this choice of $(Ha_i,Kb_j)$ is unique for a given $Kc$. Hence, $\phi$ is a bijection and $$ [G:K]=[G:H][H:K]. $$ This proof formalizes Aaron's comment about breaking each coset up into finer cosets.