Using the filter description of $\beta \mathbb N$ one can easily prove $\beta \mathbb N$ is totally disconnected; as $\mathbb N$ is a discrete space, $\beta \mathbb N$ is clearly a stone space and thus it is totally disconnected.
Now a question comes up, if $X$ is totally disconnected and Tychonoff, does this imply $\beta X$ is totally disconnected?
I tried to make a counterexample; namely, I tried to show $\beta\mathbb Q$ is not totally disconnected, but I showed that $\beta \mathbb Q$ is totally disconnected if whenever $Z_1$, $Z_2$ are disjoint closed subsets of $\mathbb Q$ there is some clopen subset $A$ of $\mathbb Q$ such that $Z_1\subseteq A$ and $A\cap Z_2=\emptyset$, but I don't know whether this last property is true or not. So, is this a counterexample?
Thanks for any help.