For reference:
A space is extremally disconnected iff the closure of every open set is clopen.
I am trying to understand a simple proof for the fact that the Čech-Stone compactification $\beta X$ of an extremally disconnected space $X$ is again extremally disconnected. The proof is:
Let $U\subseteq \beta X$ be open. Then $U \cap X$ is open in $X$ (for the subspace topology), so $\text{cl}_X(U\cap X)$ is clopen in X and $\text{cl}_{\beta X}(U) = \text{cl}_{\beta X}(\text{cl}_X(U\cap X))$ is clopen in $\beta X$. Q.E.D
My question is, why is $\text{cl}_{\beta X}(\text{cl}_X(U\cap X))$ clopen?
Maybe it is useful to point out why the last equation holds. One inclusion is clear: $$\text{cl}_{\beta X}(U) \supseteq \text{cl}_X(U\cap X) \implies \text{cl}_{\beta X}(U) \supseteq \text{cl}_{\beta X}(\text{cl}_X(U\cap X)) $$ For the other inclusion we use the density of $X$ in $\beta X$ $$\text{cl}_{\beta X}(\text{cl}_X(U\cap X)) \supseteq \text{cl}_{\beta X}(U\cap X) \supseteq U \cap \text{cl}_{\beta X}(X) = U \cap \beta X = U$$ Sadly I don't see how the clopenness of $\text{cl}_X(U\cap X)$ would imply that $\text{cl}_{\beta X}(\text{cl}_X(U\cap X))$ is clopen. All input is appreciated.