Upon discovering that denseness is transitive, I wondered if denseness is a partial ordering ($\iff$ reflexive, antisymmetric, transitive).
To be more precise: Let $X$ be a topological space. Then define $R \subset \mathcal{P}(X) \times \mathcal{P}(X)$ by $$ (a,b) \in R \iff aRb \iff a \text{ is dense in } b, $$ where $a,b \in \mathcal{P}(X)$.
Since every set is dense in itself this relation is symmetric but is it also antisymmetric ($aRb$ and $bRa \implies a = b$)? Or do I have to define it a little bit more precise as $$ (a,b) \in R \iff a \cap b \text{ dense in } b, $$ since $a$ doesn't have to be a subset of $b$?
It depends on how you define relative denseness for subsets/subspaces of $X.$ For example, we might say:
In that case, your relation $R$ is trivially antisymmetric by double-inclusion. The above is the typical definition, as Daniel Wainfleet points out in the comments.
On the other hand, you're considering that we might instead say:
In that case, we won't be able to prove antisymmetry. Consider for example $X=\Bbb R$ in the usual topology, $A=X\setminus\{0\},$ and $B=X\setminus\{1\}.$
Moreover, we won't be able to prove transitivity, either! Consider for example $A=\Bbb Q,$ $B=\Bbb R,$ and $C=\Bbb R\setminus\Bbb Q$ as subsets of $\Bbb R$ in the usual topology. Then $A$ is dense in $B$ and $B$ is dense in $C,$ but $A$ is not dense in $C.$
Consequently, it seems that the first definition is the one you want, which makes $R$ a partial order, and in fact, $R$ is a sub-relation of $[\subseteq]_{\mathcal{P}(X)}.$ You are correct that, given two sets $A$ and $B$ we don't necessarily have either of them as a subset of the other, which only means that $R$ won't be a total order (unless $X$ is empty).