If $H,K$ are subgroups of finite group $G,$ then $|G:(H \cap K)|\leq |G:H||G:K|$ 
If $H,K$ are subgroups of finite group $G,$ then $|G:(H \cap K)|\leq |G:H||G:K|$

May I know if my proof is correct? Thank you v. much. 
Proof:
$$|G:(H \cap K)| \leq |G:H||G:K| \Longleftrightarrow \frac{|G|}{|H \cap K|}\frac{|H|}{|G|} = \frac{|H|}{|H \cap K|}\leq \frac{|G|}{|K|}$$
So it suffices to prove: $|H/(H\cap K)| \leq |G/K|,$ where $G$ is not necessarily finite. Given $h \in H,$ let $$ h(H\cap K) \mapsto \ hK$$
The mapping is well-defined: 
$$h'(H \cap K) = h(H \cap K) \implies \exists m \in H \cap K: h'=hm \implies h'K = hmK= hK$$
The mapping is injective: 
$$hK = h'K \implies \exists k \in K: h=h'k \implies k \in H \cap K \implies h(H\cap K) = h'(H\cap K)$$
 A: $\renewcommand{\phi}{\varphi}$I think it is preferable to prove first a result of general interest, that is
$$
\lvert H K : K \rvert = \lvert H : H \cap K \rvert.
$$
Here $HK$ is not necessarily a subgroup, but it is definitely a union of right cosets $h K$ of $K$, for $h \in H$. 
This I believe you have implicitly proved, anyway it goes as follows. Compose the map $H \to H K$ given by $h \mapsto h \cdot 1$ with the map $HK \to  HK/K$ given by $g \mapsto gK$ to get the map $\phi: H \mapsto HK/K$ given by $h \mapsto h K$. This is clearly onto, and for $a, b \in H$ one has 
\begin{align}
\phi(a) = \phi(b)
\text{ iff }& 
a K = b K
\\\text{iff } &
a^{-1} b \in H \cap K
\\\text{iff }&
a (H \cap K) = b (H \cap K).
\end{align}
Thus $\phi$ induces a bijection $H/H \cap K \to HK/K$.
Now argue as you did
$$
\lvert G : H \cap K \rvert
=
\lvert G : H \rvert \cdot \lvert H : H \cap K \rvert
=
\lvert G : H \rvert \cdot \lvert HK : K \rvert
\le
\lvert G : H \rvert \cdot \lvert G : K \rvert.
$$
A: A simpler solution is to prove that the map
$
\qquad G/(H\cap K) \to G/H \times G/K$
given by
$
g(H\cap K) \mapsto (gH, gK)
$
is well defined and injective.
(The quotients here are sets of cosets.)
