$\beta \omega$ is zero dimensional and extremally disconnected I have proved that $\beta \omega$ is zero dimensional by constructing it using ultrafilters, but I know that $\beta \omega$ is characterized up to topological equivalence by the extension property, so I would like to know if there is a way to prove it using the extension property.
Also, I'm looking for a simple proof for "$\beta \omega$ is extremally disconnected".
 A: We can use that $\beta D$, where $D$ is discrete, is a projective space in the category of compact Hausdorff spaces. A compact Hausdorff space $X$ is a projective space in the category of Hausdorff spaces (for short projective) iff for all compact Hausdorff spaces $Y$ and $Z$, every continuous and surjective $f: Y \rightarrow Z$ and every continuous $g: X \rightarrow Z$, there exists a continuous $h: X \rightarrow Y$ such that $g = f \circ h$. 
Theorem: a projective space $X$ is extremely disconnected, i.e. the closure $\overline{G}$ of an open set $G$ of $X$ is open.
Proof: Let $G \subset X$ be open. We consider $X \times \{0,1\}$, where the space $\{0,1\}$ has the discrete topology (this can also be seen as a disjoint sum of two copies of $X$), which is also compact Hausdorff. Define $Y = (X\setminus G) \times \{0\} \cup \overline{G} \times \{1\} \subset X \times \{0,1\}$, which is compact as well, and note that the function $f: Y \rightarrow X$, with $f(x,i) = x$ for $i=0,1$, $x \in X$, is continuous (just the restricted projection onto $X$) and onto. So we can apply the definition of projectiveness to $f$ and $g = 1_X$, the identity on $X$, to get a continuous $h: X \rightarrow Y$ such that $f \circ h = 1_X$.
For $x \in G$, $h(x) = (y,i)$ for some $(y,i)$ in $Y$, but as $f(h(x)) = f((y,i)) = y$ which should equal $x$, we know that $h(x) = (x, i)$. But by the definition of $Y$, this means that $i = 1$. So for all $x \in G$, $h(x) = (x,1)$. But by continuity of $h$ and closedness of $\overline{G} \times \{1\}$ in $Y$, we know that $h(x) = (x,1)$ for all $x \in \overline{G}$ as well. But this means that $\overline{G} = h^{-1}[\overline{G} \times \{1\}]$, and the latter set is the inverse image of an open set in $Y$ (both parts of $Y$ are clopen in $Y$!) so indeed open in $X$, as required. This completes the proof. 
Theorem 2: For any discrete space $D$, the space $\beta D$ is projective.
Proof: let $Y,Z$ be compact Hausdorff, and $f: Y \rightarrow Z$ be onto continuous and $g: \beta D \rightarrow Z$ be continuous. First define $h$ on $D$: for $d \in D \subset \beta D$, we pick any point in $f^{-1}[\{g(d)\}]$, which we call $h(d)$. This can be done by the fact that $f$ is surjective and using the axiom of choice (which we need anyway to get $\beta D$ to exist...). By definition, $f \circ h = g$ holds on $D$. But by the universal property we have a continuous $\beta h : \beta D \rightarrow Y$ that extends $h$. By continuity of all maps involved, the identity also holds on $\overline{D} = \beta D$, as required.
Corollary: $\beta D$ is extremally disconnected for all discrete $D$.
Theorem 3: If $X$ is extremely disconnected and regular, $X$ is zero-dimensional.
Proof: if $O$ is open and $x \in O$, find $U$ open such that $x \in U \overline{U} \subset O$ by regularity. We also know that $\overline{U}$ is open, by $X$ being extremely disconnected. So that set is a clopen subset containing $x$ contained in $O$. So the clopen sets form an open base, so $X$ is zero-dimensional.
Corollary 2: $\beta D$ is zero-dimensional for discrete spaces $D$. 
Easy, as a compact Hausdorff is regular.
This provides a proof without using ultrafilters, but notions using the universal property, so more category theory-like in spirit. 
