$\mathbf{A.}$ Define a partial order (Domain,<,PO) as follows:
PosNeg := {1,-1, 2,-2, 3,-3 ....}
Domain := {FinSub$_i$ : Finsub$_i$ is a finite subset of PosNeg}
PO is ordered using "<" with the following rules:
FinSub$_i$ < FinSub$_j$ $\;$ iff $\;$ FinSub$_i$ $\subset$ FinSub$_j$ ...............................(1)
$\mathbf{B.}$ Define a Cohen Dense "D" subset of Domain having the following two properties :
(i) $\forall$ x $\in$ Domain $\exists$ y $\in$ D (x < y) .......................(2)
(ii) IF [(x $\in$ D) AND (x < y) AND (y $\in$ Domain) ] THEN y $\in$ D.......................(3)
$\mathbf{C.}$ $\mathbf{My}$ $\mathbf{Question}$ $\mathbf{is:}$
How to show that any 'consistent and complete' infinite subset "CC" of PosNeg will intersect every Cohen Dense set D (consistent means : if i/-i is in CC then -i/i is not in CC, and 'complete' means for every i, then i or -i is in CC)?
(Note this question is an attempt to 'modernise' the question Cohen Forcing in Set Theory - Proof that Forcing is Equivalent to intersection of Dense Sets).
My attempt at this proof showed that a Cohen Dense set "PerverseD" could always be chosen that would $\mathbf{stop}$ CC intersecting "PerverseD", making the question statement false (but Cohen indicates it is true).
My construction of this "PerverseD" is constructed by applying (2) first then (3), e.g. if CC:={1,2,...,n,.....} with all positive numbers:
If x={1,2,3,....,n} put {1,2,3,...,n,-(n+1)} $\in$ "PerverseD". Then take all possible supersets of these sets.
So as n tends to infinity there will be always be a finite part of CC that can be a subset of a set in "PerverseD", but none are ever exactly the same, for finite n.
In fact, no maximal filter (= complete consistent set) in any nontrivial forcing notion (you're looking at Cohen forcing in particular) meets every dense set: if $\mathbb{P}$ is a forcing notion and $G$ is a filter in $\mathbb{P}$ the set $\mathbb{P}\setminus G$ is a dense (by nontriviality) subset of $\mathbb{P}$ which $G$ does not meet.
(In your language: if $G$ is any complete consistent set, the collection of elements of the partial order which "disagree" with $G$ at some point is a dense set which $G$ doesn't meet.)
In particular, if $M$ is a model of ZFC and $\mathbb{P}$ is a nontrivial forcing notion in $M$, no filter which is $\mathbb{P}$-generic over $M$ will exist in $M$. Of course if $M$ is countable there will indeed be $\mathbb{P}$-generic filters of $M$, but they won't be elements of $M$.
What happens if we try to run the argument above inside a countable $M$? Well, we need to argue that for every filter $G$ the dense set $\mathbb{P}\setminus G$ is actually in $M$. But there's no reason to believe that it is. In fact, it never will be, since that would contradict the fact that generics over $M$ do exist.
And here's a proof of that: since $M$ is countable, the set $\mathcal{D}$ of dense subsets of Cohen forcing which are elements of $M$ is countable. We know that for any countable set of dense sets there is a filter meeting them all.