Basics on group operation Let's $G$ be a group and $H \le G$. $H$ brings a partition of $G$, $\lambda_H(G)=(H,\complement_GH$), where $\complement_GH:=G \setminus H$. Firstly, I notice that, as $g \in G \Leftrightarrow g^{-1} \in G$, then also $g \in \complement_GH \Leftrightarrow g^{-1} \in \complement_GH$.

Now, I claim that $g \in H, g' \in \complement_GH \Rightarrow gg' \in \complement_GH$. In fact, $gg' \in H \Rightarrow g'=g^{-1}(gg') \in H$: contradiction. Then, $gg' \in \complement_GH$.

So, denoted by $f \colon G \times G \rightarrow G$ the group operation, we get:


$f(G \times G) \subseteq G$
$f(H \times H) \subseteq H$
$f(H \times \complement_GH) \subseteq \complement_GH$
$f(\complement_GH \times H) \subseteq \complement_GH$


Finally, I claim that $f(\complement_GH \times \complement_GH) \cap H \ne \emptyset$. In fact, differently, we'd have that $f(\complement_GH \times \complement_GH) \subseteq \complement_GH$ and then $\complement_GH \le G$: contradiction, because $e \notin \complement_GH$.

Is this whole correct?
It is not really clear what you are asking about. Though, your reasoning is correct.

But, note that, you cannot say that $H$ partitions the group $G$. It would be saying that it is a set's property to partition any other sets/groups/categories etc. in which it is included. You see, having a complement is not what we call partitioning, even if it seems similar. With complements, it is already too trivial! For example, you could omit all your proof between second line and the before last one: As $H$ is a subgroup of $G$, thus the identity element of $G$ is included in $H$, thus no way it can be for the complement of $H$; so that we conclude the complement can never be a subgroup of $G$.

But the properties you showed would still remain valid.

Except that: 


When you tried to proceed by contradiction at the last line, you began with a logically false claim. The proof holds on itself. But it is not how it is done. We prefer(!) beginning with a logically valid statement which doesn't hold with the given propositional phrase, then we try to arrive either to a logically false conclusion or any conclusion which does not hold with predicate to prove a contradiction.
I surely think that $ f( \bar{H} \times \bar{H}) \subseteq \bar{H} $ cannot imply that $\bar{H}$ is a subgroup of $G$ too. (where $\bar{H}$ is the complement). It only shows an inclusion; and nothing about a property at all.
The inclusion relation is what we call an ordering. It is, in fact, a linear ordering in this case. A linear ordering is a partial ordering in which the comparaison quality is well defined for every element pairwise distinct. Now, I will initiate you to a higher theorem: 

("higher" in the sense that, here, we don't make use of it within its usual definition for limits; but for a different metric, ie. topological space with different ordering, but which holds well because they are isomorphic, ie. homoemorphic.)

So, you can use here the Sandwich Theorem. You see, the image of complements is ordered by usual set inclusion between the complement of $G$ and the $H$ itself. Though we know that we cannot have any element outside $G$. But we cannot have any in $H$ neither: Otherwise we would have images of $H$ and its complement intertwining. And as $f$ is a homomorphism of groups, it cannot be, thus contradiction.

You see, with the reasoning I told above, adapting it to different cases is only up to your redaction and comprehension. It becomes free of your choice for underlying set (ie. $Dom(f)$ here). You just need: 1. the linear ordering of inclusion (which you surely have here -unless in higher mathematics), 2. the function $f$ to be homomorphism of groups.