Filter without cluster point, then the clopen members have empty intersection Consider a topological space $(X,\tau)$ and a filter $F$ on $X$ with no cluster point. The set $C$ of all clopen members of $F$ has the finite intersection property. Why has the intersection $\bigcap_{x \in C} x$ to be empty?
I cannot find a way to show that the fact, that the intersection is not-empty implies that every element of the filter $N$ of neighbourhoods of $x$, has a non-empty intersection with every element of $F$.
If this would be the case, then $N \cup F$ would yield a subbasis for a convergent filter: a contradiction to the fact, that $F$ has no cluster point.

EDIT
It seems, that I generalized the problem too much. Take a look at the following proof of Herrlich's Axiom of Choice (Theorem 4.92 about equivalence of Ascoli Theorem w.r.t. ultrafilter=compactness and PIT = Boolean Prime Ideal Theorem):

In $(1)\Rightarrow(2)$, the fact, that $P=\mathfrak{2}^I$ is not compact w.r.t. the open covering property leads to a filter $F$ with no cluster point.
What am I missing?
 A: Why would this be true, without further assumptions? Take a filter on the reals (usual topology) without a cluster point (which can be done as the reals are not compact). The only clopen member of any filter is $\mathbb{R}$ itself, by connectedness, and this has non-empty intersection... 
A: Your original question was already answered by Henno Brandsma. I will try to respond to the new version of your question.
The key here is probably the fact that $2^I$ is zero-dimensional. I.e., it has a base consisting of clopen sets. So let us check whether the claim holds in such spaces.
Suppose that $\mathcal F\subseteq \mathcal P(Z)$ is a filter and $Z$ is a zero-dimensional space. Suppose that $\mathcal F$ has no cluster point, meaning that $$\bigcap_{F\in\mathcal F}\overline F=\emptyset.$$
Now let $z\in Z$. Since $z$ is not a cluster point of $\mathcal F$, there exists and $F\in\mathcal F$ such that $z\notin\overline F$. Then 
there exists a (basic) clopen neighborhood $U$ of the point $z$ 
$$U\cap F=\emptyset.$$
From this we get that $Z\setminus U\supseteq F$ and
$$Z\setminus U\in\mathcal F.$$
So we found a clopen set in the filter $\mathcal F$ which does not contain the point $z$. This implies that
$$z\notin\bigcap\{A\in\mathcal F; A\text{ is clopen}\}.$$
Since this is true for every $z\in Z$, we get that
$$\bigcap\{A\in\mathcal F; A\text{ is clopen}\}=\emptyset.$$
