Proof. ($\Rightarrow$): Suppose that $X$ is extremely disconnected. Let $\cf \subset C(X; \real)$ and $F \in C(X; \real)$ be an upper bound of $\cf$. For each $q \in \real$, let

then $U(q)$ is open. Now, define

For each $x \in X$, $g(x) \ge \sup_{f \in \cf}f(x)$, so $g(x) > -\infty$. On the other hand, for any $q > F(x)$, $x\not\in \overline{\bracs{F > q}}$, so $g(x) < \infty$. Thus $g$ is defined as a real-valued function.

Let $p \in \real$, then $\bracs{g > p}= \bigcup_{q \in \real, q > p}U(q)$ is open. On the other hand,

which is also open, so $g$ is continuous.

Now, for each $p \in \real$, since $F$ is continuous,

so $g \le F$. As this holds for all upper bounds of $\cf$, $g$ is the supremum of $\cf$ in $C(X; \real)$.

($\Leftarrow$, [Theorem II.9.6, Zhu93]): Suppose that $X$ is completely regular and $C(X; \real)$ is order complete. Let $U \subset X$ be open and

and $F$ be the supremum of $\cf$ in $C(X; \real)$. Since $X$ is completely regular, for each $x \in U$, there exists $f \in C(X; [0, 1])$ with $f(x) = 1$ and $f|_{U^c}= 0$. Thus $f \in \cf$ and $\one_{\bracs{x}}\le f \le F$. As this holds for all $x \in U$, $F \ge \one_{U}$.

On the other hand, for any $x \in \ol{U}^{c}$, there exists $g \in C(X;[0, 1])$ with $g(x) = 0$ and $g|_{\ol{U}}= 1$. In this case, $g \ge f$ for all $f \in \cf$, so $g \ge F$ as well. This yields that $F \le g \le \one_{X \setminus \bracs{x}}$. As this holds for all $x \in X \setminus \ol{U}$, $F \le \one_{\ol{U}}$.

Given that $\one_{U} \le F \le \one_{\ol{U}}$ and $F$ is continuous, $\bracs{F = 1}\supset \ol{U}$, so $F = \one_{\ol{U}}$ is continuous. Hence $\ol{U}= F^{-1}(1)$ must be open.$\square$