Proof Verification of Coset Multiplication Please verify my following proof:
Theorem: Let $H$ be a subgroup of $G$, then the left coset multiplication is well defined by the equation: $(aH)(bH)=(ab)H$ iff $H$ is a normal subgroup. 
Proof:
$\boxed{\rightarrow}$  $(aH)(bH)=(ab)H$ as coset multiplication is well defined. Then assigning $ah_1$ and $bh_2$ as representatives to $aH$ and $bH$ respectively we have (Italic is added for edditing as per comment suggestion):  $$(aH)(bH) \rightarrow (ah_1)(bh_2) = (ab)h_3  \rightarrow h_1bh_2 = bh_3 \rightarrow h_1b=bh_3h_2^{-1} \rightarrow h_1b=bh_4$$ 
So all elements of $bH$ can be written in the form $Hb$, so $Hb=bH$, so $H$ is normal. 
$\boxed{\leftarrow}$ $H$ is normal, so $aH=Ha$, need to show that $(aH)(bH)=(ab)H$. Then assigning $ah_1$ and $bh_2$ as representatives to $aH$ and $bH$ respectively we have (Italic is added for edditing as per comment suggestion):
$$(aH)(bH) \rightarrow (ah_1)(bh_2) = a(h_1b)h_2 = a(bh_3)h_2 = ab(h_3h_2) \rightarrow (ab)H$$ closure as $h_2$ and$h_3$ are arbitrary elements of the subgroup $H$.
QED. 
I know that double inclusion is standard for proving set equality, but if I can show that all elements of set A can be written as all elements in set B, does that suffice to show $A=B$
Explicitly, I  showed that all elements in $aH$ can be expressed as elements of $Ha$. Does that suffice to show $aH=Ha$
 A: Your proof is correct.
Mybe you can give some little explanations in the direction $\longrightarrow,$ but
in general is o.k.
A: Your proof is good. And yes, showing $aH \subseteq Ha$ is not enough, but usually there is some sort of symmetry: your argument works the other way round perfectly.

I do think the proofs can be streamlined a little. Firstly, note

$H$ is normal iff for all $a \in G$, $aHa^{-1} \subseteq H$.

This is an equivalent definition: it implies $aHa^{-1} \subseteq H$ and $a^{-1}Ha \subseteq H$. So $aH \subseteq Ha$ and $Ha \subseteq aH$ giving $aH=Ha$. The only advantage of this is we do not have to check equality.
Proofs:

$(\Rightarrow)$ We have $(aH)(a^{-1}H) = H \Rightarrow (aHa^{-1})H = H \Rightarrow aHa^{-1} \subseteq H$.
$(\Leftarrow)$ As $H$ is normal, $Hb=bH$, so $(aH)(bH) = a(Hb)H=a(bH)H=abH$, using $HH=H$.

The fact that I can manipulate the way I did in second line follows from associativity of group elements. For example, $(ah_1)(bh_2)= a(h_1b)h_2$.

I hope this helps a little.
