Intersection of cosets from possibly distinct subgroups is either empty or a coset of the intersection between the two subgroups 
Let  $G$ be a group and $H \leq G, K \leq G$.
The intersection of two left cosets (one from $H$ and the other from $K$) is either empty or a coset of
$H\cap K$

I'm having some trouble proving this in a good way.
We know that, if $G$ is a group and $H \leq G, K \leq G$ then $(H\cap K) \leq G$.
Let us a consider $H\cap K$ and some coset $a(H\cap K)$ of $H\cap K$ for a arbitrary $a\in G$
If $x \in a(H\cap K)  \implies x =ab,$ for some $b \in H\cap K \implies x \in aH$ and $x \in aK \implies x \in aH\cap aK $. But this is not what I want, right? Or can this implication be done the other way as well?
One thought is to split it up in to cases.

*

*$H\cap K = \{e \}$, where e is the identity.

*$H = K$ or $H \subset K$ or $K \subset H$ or $H\cap K \neq \{e \}$.
But I'm not sure if this would get me anywhere?

If I want to prove that the intersection must be empty I'm not sure where to start. I want to assume that some $x\in G$ is contained in the intersection of two cosets and then perhaps derive that it must be empty.
Should I then take $x \in aH\cap aK $ or $x \in aH\cap bK $ where $a,b$ may or may not be distinct?
If I take $x \in aH\cap aK $ then $x = ah, h\in H$ and $x = ak, k\in K \implies x = ah = ak \implies h = k$. And I'm stuck.
If I take $x \in aH\cap bK $ then $x = ah, h\in H$ and $x = bk, k\in K \implies x = ah = bk \implies a^{-1}bk \in H$ And I'm also stuck here.
Any hints on how to solve this?

dani_s's answer has a typo in the second paragraph. It should state $a=xh^{-1}, b=xk^{-1}$ instead ($h=xa^{-1}$ would be wrong without assuming abelian group).
Also the middle section of the proof can be shortened:
$$aH=xh^{-1}H=x(h^{-1}H)=xH \quad\text{ since }h^{-1} \in H$$
and
$$bK=xk^{-1}K=x(k^{-1}K)=xK \quad\text{ since }k^{-1} \in K.$$
