I'm finally struggling a little bit with the notion of forcing, so I decided to do some of the exercises in Jech's book (2nd Edition) in order to sharpen a little bit my intuitions about the whole thing. Anyway, I'm stuck a little bit on exercise 16.5 (14.5 in the new edition, I believe), so I'd appreciate some help here. First, however, some definitions (I'll omit the references to the ground model $M$ here, as I don't think they're relevant to the exercise at hand; basically, everything except the generic filter is assumed to be in $M$).
Let $\langle P, \leq \rangle$ be a partial order. If $p, q \in P$ are such that there is an $r \in P$ and both $r \leq p$ and $r \leq q$, we say that $p$ and $q$ are compatible; otherwise, they are incompatible. A cover $C$ of $P$ is a set such that for every $p \in P$, there is a $q \in C$ such that $p$ and $q$ are compatible. A a set $F \subseteq P$ is a filter if:
$F \not = \emptyset$;
if $p \in F$ and $p \leq q$, then $q \in F$;
if $p, q \in F$, then there is an $r \in F$ such that $r \leq p$ and $r \leq q$.
A filter $G$ on $P$ is called generic iff $G \cap C \not = \emptyset$ whenever $C$ is a cover of $P$.
Now define a partition $W$ of $P$ as a subset of $P$ such that $W$ is a cover of $P$ and for any distinct $p, q \in W$, $p$ and $q$ are incompatible. I'm asked to prove: a filter $G$ on $P$ is generic iff $G \cap W \not = \emptyset$ whenever $W$ is a partition of $P$. From left to right is immediate from the definition of generic: if $G \cap C \not = \emptyset$ whenever $C$ is a cover of $P$, then a fortiori $G \cap W \not = \emptyset$ whenever $W$ is a partition of $P$, as every partition of $P$ is, by definition, a cover of $P$. The side that is bugging me is from right to left. Here's what I got so far:
Let $p \in G \cap W$ and suppose $C$ is an arbitrary cover of $P$. Since $C$ is a cover of $P$, there is a $q \in C$ such that $p$ and $q$ are compatible, i.e. there is an $r$ such that $r \leq p$ and $r \leq q$. But $W$ is also a cover of $P$, so there must be an $s \in W$ such that $s$ and $q$ are compatible, i.e. there's a $t$ such that $t \leq s$ and $t \leq q$. Now, suppose $q < p$. Since $t \leq q$, by transitivity it follows that $t < p$, whence $p$ and $s$ are compatible. But that contradicts the fact that $p, s \in W$. So it's impossible for $q < p$.
From the above, I'd like to conclude that $p \leq q$, so that $q \in G$ and, hence, $G \cap C \not = \emptyset$, which would finish the proof. However, I only have a partial order, not a total order. Thus, it could be the case that $p$ and $q$ are incomparable, blocking the conclusion. So I'm beginning to think that my strategy may be wrong. Is there anything I'm missing?
HINT: Suppose that $C$ is a cover, well-order it as $\{p_\alpha\mid\alpha<\lambda\}$ and by transfinite recursion, define a partition.
(The axiom of choice is essential, since without the axiom of choice, it might be that a partial order has no partitions, so every filter is vacuously generic by the second definition.)