Lemma 2.20 of Kunen

I'm trying to understand the proof of a result in Kunen's Set Theory An Introduction to Independence Proofs Chapter VII, but I really can't find a way.

We want to show that if $G$ is a $\mathbb{P}$-generic filter then either $G \cap E \neq \emptyset$ or there exists $q \in G$ such that for all $r \in E$ we have that $q \perp r$ meaning that there is no common extension p such that $p \leq q$ and $p \leq r$.

We are asked to consider the dense set $D= \lbrace p \mid \exists r \in E \,\, (p \leq r) \rbrace \cup \lbrace q \mid \forall r \in E \, (r \perp q) \rbrace$ and in the end we are told that $G\cap D \neq \emptyset $ implies our claim. But I can't see why this last implication should hold, and in particular why the intersection could not be included in $ \lbrace p \mid \exists r \in E \,\, (p \leq r) \rbrace \ \setminus E$. Maybe this would contradict the filter properties, but it seems that a filter as defined by Kunen (see page 53) is closed under taking elements that are "LESS THAN", not "GREATER THAN".

Take any $p\in G\cap D$. Either $\exists r\in E$ such that $p\leq r$, in which case since $G$ is a filter, $r\in G$, so $G\cap E\neq\varnothing$. Or else $\forall r\in E$ we have that $r\perp p$. But now if $r\in G\cap E$, then because $G$ is a filter, $r$ and $p$ are compatible, which is impossible.