My version of the theorem is the following: Let $\{E_n:n\in\mathbb{N}\}$ be a countable family of subsets of a Boolean algebra, $\mathbb{A}$, having infs and sups. Show that there is an ultrafilter $Q$ which is $E_n$-complete for all $n\in \mathbb{N}$.
My attempt at a proof: Let $\{E_n:n\in \mathbb{N}\}$ be given as above. Let $a>0$ be fixed, by previous work I have that $$d(E_n) =\{a>\mathbb{O}:\forall b\in E_n (a\cdot b=\mathbb{O})\}\cup\{a>\mathbb{O}:\exists b\in E_n (a\le b)\}$$
is dense. We have that $d(E_1)$ is dense so there exists $b_1\in d(E_1)$ such that $b_1\leq a$. Next, for all $n>1$, we have that $d(E_n)$ is dense, so there exists $b_n \in d(E_n)$ such that $b_n \leq b_{n-1}$. We thus inductively obtain a decreasing sequence of $b_n$s with $b_n \in d(E_n)$.
Let $F_{b_n}=\{x: b_n\le x\}$, then let $Q=\bigcup_{n\in\mathbb{N}} F_{b_n}$.
Suppose $E_n \subseteq Q$, then $E_n \subset F_{b_m}$ for some $m$. Thus, for each $e \in E_n$, $b_m \leq e$, so $b_m \cdot e = b_m$ we thus have that $\inf \{b_m \cdot e : e \in E_n\} = \inf \{ b_m \} = b_m$. But, by Exercise 2.6, this is $\inf \{b_m \cdot e : e \in E_n\} = b_m \cdot \inf \{e : e \in E_n\}$, so we have that $b_m \cdot \inf \{e : e \in E_n\} = b_m$, that is, $b_m \leq \inf \{e : e \in E_n\}$, so $\inf \{e : e \in E_n\} \in F_{b_m} \subset Q$. $Q$ is thus $E_n$-complete.
The only thing I don't know is how to show that $Q$ is an ultrafilter.... I tried a lot of things but I don't see it. Hints please?
Your $Q$ need not be an ultrafilter, but it is a filter and can be extended to an ultrafilter, which I’ll call $U$. It’s $U$ that you need to show is $E_n$-complete for each $n$. (And even if your $Q$ were an ultrafilter, I don’t see how you conclude that $E_n\subseteq F_{b_m}$ for some $m$ if $E_n\subseteq Q$.)
To show that $U$ is $E_n$-complete I find it easier to use the alternative characterization of $E_n$-completeness: if $\sup E_n\in U$, then $E_n\cap U\ne\varnothing$. Let $t_n=\sup E_n$, and suppose that $t_n\in U$; $b_n\in U$, so $t_n\cdot b_n\in U$. In particular, $t_n\cdot b_n\ne\Bbb O$, so, since $b_n\in d(E_n)$, there must be an $e_n\in E_n$ such that $b_n\le e_n$. But then $t_n\cdot b_n\le t_n\cdot e_n=e_n$, so $e_n\in U$, and therefore $e_n\in E_n\cap U\ne\varnothing$, as desired.