Basically disconnected space which is not extremally disconnected 
Space $X$ is basically disconnected if every cozero-set has an open closure.

Every extremally disconnected space is basically disconnected But i think the converse fails. The one-point compactification $\alpha D(\tau)$ of an uncountable discrete space is a counterexample? 
 A: That doesn’t quite work: if $C$ is a countably infinite subset of $D(\tau)$, $C$ is a cozero-set in $\alpha D(\tau)$ whose closure isn’t open.
Take $X$ to be the one-point Lindelöfization of an uncountable discrete space, with non-isolated point $p$: nbhds of $p$ are the sets $X\setminus C$ for countable $C\subseteq(X\setminus\{p\})$. $X$ is a $P$-space, meaning that $G_\delta$-sets in $X$ are open, so every zero-set in $X$ is open, and therefore every cozero-set in $X$ is closed, and $X$ is basically disconnected.
Now let $X\setminus\{p\}=U\cup V$, where $U$ and $V$ are uncountable and disjoint. $U$ and $V$ are open, but $\operatorname{cl}U=U\cup\{p\}$ and $\operatorname{cl}V=V\cup\{p\}$ are not disjoint, so $X$ is not extremally disconnected.
A: One can use Stone duality to construct many counterexamples. It is not to hard to show that a Boolean algebra $B$ is complete if and only if the Stone space $S(B)$ is extremally disconnected. Furthermore, a Boolean algebra $B$ is $\sigma$-complete (recall that a Boolean algebra is $\sigma$-complete iff every countable subset has a least upper bound) if and only if the Stone space $S(B)$ is basically disconnected. Therefore, if $B$ is a $\sigma$-complete but not complete, then $S(B)$ is basically disconnected but not extremally disconnected. Moreover, a zero-dimensional space is extremally disconnected if and only if the Boolean algebra of clopen sets is a complete Boolean algebra. One can easily show that your proposed counterexample is not a true counterexample by looking at that Boolean algebra of all clopen sets.
A: you can find examples in the Gillman and Jerison's book: Ring of continuous functions
