Suppose we do have a filter $\mathcal{F}$ on $\omega$ which contains the cofinite filter, so $X\in\mathcal{F}$ implies $X$ is infinite. For $X\in\mathcal{F}$, let $f_X$ be the increasing enumeration of $X$. Let $\tilde{\mathcal{F}}=\{f_X\mid X\in\mathcal{F}\}$. Let $f\in\omega^\omega$ s.t. $f$ is a bound of $\tilde{\mathcal{F}}$.
Identify $X\subseteq\omega$ with $x\in2^\omega$ defined by $x(n)=1$ if $n\in X$ and $x(n)=0$, if $n\notin X$ and consider the discrete topology on $2$ and the product topology in $2^\omega$.
I want to prove, that $A_n:=\{X\subseteq\omega\mid\forall k\geq n\ f_X(k)\leq f(k)\}$ is meager for all $n\in\omega$. Any hints?
Definition of an unbounded familiy $F$: A family of subsets of $\omega^\omega$ is unbounded, if for all $g\in \omega^\omega$ exists a $f\in F$ s.t. $\neg(f\leq^* g)$, where $f\leq^*g$ means $f(n)\leq g(n)$ for all but finitely many $n\in\omega$.
First, note that $\mathcal{F}$ is a red herring; the real question is
Note that I've also gotten rid of $n$; this is just for simplicity, and won't change anything important.
EXERCISE: for every $\sigma\in 2^{<\omega}$ there is some $\tau\in 2^{<\omega}$ with $\sigma\preccurlyeq\tau$ such that for any $X$, if the characteristic function of $X$ extends $\tau$, then $X$ is not in $A_f$.
If you can do this, do you see why this means $A_f$ is in fact nowhere dense (not just meager)? (The passage to meager comes when we bring $n$ back into the picture.)
By the way, this line of attack comes from the game picture of category; see e.g. my answer to Analogy for Baire categories?.