Consider $R=\Pi_{i\in \mathbb{N}} k$, where $k$ is a field. We know that its Spectrum $Spec(R)$ is quasi-compact. Moreover, by many answers already on this site, we know that the Krull dimension of $R$ is $0$ and if you use ultrafilters you can prove that this is the Stone-Cech compactification of $\mathbb{Z}_+$.
Still, I have the following twist in my mind: Consider the element $e_i$, which has $1$ in every copy of $k$ except in the $i$-th, where it is $0$. Then $e_i$ is a non-trivial idempotent. Furthermore we observe $V(e_i)\cup V(1-e_i) = Spec(R)$ and $V(e_i) \cap V(1-e_i) = \emptyset$. Since $e_i$ and $1-e_i$ are not units, $V(e_i)$ and $V(1-e_i)$ are both non-empty. Moreover for $i \neq j$ we have that $V(e_i) \cap V(e_j) =V((e_i, e_j))=\emptyset$, since the latter ideal contains the unit $e_i+e_j$. Thus we have constructed infinitely many disjoint subsets, which are open and closed.
As already reasoned above, there has to be a mistake in the above argument. Somehow I am not able to find it.