Answer by Gro-Tsen for Stone-Čech boundary is not extremally disconnected

The question has already been answered satisfactorily, but I think the following presentation, which is implicit in the second part of YCor's answer, will clarify things. The fact that $\beta\mathbb{N} \setminus \mathbb{N}$ isn't extremally disconnected follows from three observations which are each interesting in their own right:

1. The Boolean algebra $\operatorname{Clop}(X)$ of clopen subsets of an extremally disconnected space $X$ is complete. Indeed, the infinite join of a family $(F_i)_{i\in I}$ of clopen sets is explicitly given by $\bigvee_{i\in I} F_i = \operatorname{closure}(\bigcup_{i\in I} F_i)$ (that the space is extremally disconnected ensures that this closure is, in fact, clopen, and it is clearly the smallest clopen set containing all the $F_i$). Actually, we have a sort of converse: the Stone space of a complete Boolean algebra is extremally disconnected (such spaces are called “Stonean”).

2. The Boolean algebra $\operatorname{Clop}(\mathbb{N}^*)$ of clopen subsets of $\mathbb{N}^* := \beta\mathbb{N} \setminus \mathbb{N}$ is that $\mathscr{P}(\mathbb{N})/\textrm{fin}$ of subsets of $\mathbb{N}$ modulo finite differences. (In fact, $\mathbb{N}^*$ is the Stone space of $\mathscr{P}(\mathbb{N})/\textrm{fin}$.) This is explained in a comment to YCor's answer for the particular situation at hand; but more generally: if $X$ is a Stone space, and $Y\subseteq X$ is closed, then $\operatorname{Clop}(Y) = \{F\cap Y : F\in \operatorname{Clop}(X)\}$: see Monk & Bonnet eds. (Koppelberg), Handbook of Boolean Algebras (1989), chap. 3, lemma 7.6(b); so $F \mapsto F\cap Y$ defines a surjective homomorphism of Boolean algebras $\operatorname{Clop}(X) \to \operatorname{Clop}(Y)$, so that $\operatorname{Clop}(Y)$ can be seen as the quotient of $\operatorname{Clop}(X)$ by the equivalence relation of “coinciding on $Y$”, and in the special case of $X=\beta\mathbb{N}$ and $Y=\mathbb{N}^*$ it is clear that the kernel of $\mathscr{P}(\mathbb{N}) \to \operatorname{Clop}(\mathbb{N}^*)$ consists of finite subsets of $\mathbb{N}$ because they are the ones that belong to no free ultrafilter. (Side note: analogously, the ring $C^*(Y)$ of bounded continuous real-valued functions on $Y$ consists of restrictions to $Y$ of the functions in $C^*(X)$ (see, e.g., Gilman & Jerrison, Rings of Continuous Functions, problem 3D), so that here $C^*(\mathbb{N}^*)$ consists of bounded sequences of real numbers up to finite differences. The Boolean algebras under discussion are the Boolean algebras of idempotents of these rings of bounded continuous real-valued functions.)

3. The Boolean algebra $\mathscr{P}(\mathbb{N})/\textrm{fin}$ of subsets of $\mathbb{N}$ modulo finite differences is not complete. Here I can only repeat YCor's argument: if $\mathbb{N}$ is partitioned into countably many infinite sets $S_n$, and if $Z$ represents their putative upper bound $[Z]$ in $\mathscr{P}(\mathbb{N})/\textrm{fin}$, then we can pick an element $z_n \in Z \cap S_n$ (because each $S_n \setminus Z$ is finite) for each $n$ and let $Z' := Z \setminus \{z_n : n\in\mathbb{N}\}$: then $[Z'] < [Z]$ in $\mathscr{P}(\mathbb{N})/\textrm{fin}$ yet $[S_n] \leq [Z']$ for each $n$ since each $S_n \setminus Z'$ is finite, and this contradicts the assumption that $[Z]$ is the upper bound of the $S_n$. (Here I write $[P] \in \mathscr{P}(\mathbb{N})/\textrm{fin}$ for the class mod finite sets of an element $P \in \mathscr{P}(\mathbb{N})$.)

Putting these three facts together we get the desired conclusion, and, again, the proof is precisely that given in the second part of YCor's answer, but I think the above helps clarify what is going on.