Recall that an element $a$ of a Boolean algebra $B$ is called quasi-compact if whenever $a = \bigvee_{i \in I} a_i$ then there is finite $I_0 \subseteq I$ such that $a = \bigvee_{i \in I_0} a_i$. In the proof of Theorem 3.4.3 in Johnstone's Sketches of an Elephant the following claim is made (he uses the terminology compact instead of quasi-compact).
Claim. A Boolean algebra whose top element is quasi-compact is finite.
I cannot think of an argument for this claim. However, the following is not hard to prove.
Proposition. A Boolean algebra $B$ is finite if and only if it is complete and its top element is quasi-compact.
Proof. Left to right is trivial. For the converse we assume that $B$ is infinite and we inductively find non-zero $a_0, a_1, \ldots$ so that $a_i \wedge a_j = 0$ for all $i \neq j$ by picking $a_i$ such that there are infinitely many elements $b$ with $b \leq \neg \bigvee_{j \leq i} a_j$. Set $a = \bigvee_{i \in I} a_i$ (this is where we use completeness) and note that any element of $B$ is quasi-compact because the top element is quasi-compact. So $a = \bigvee_{i < n} a_i$ for some $n$ and we arrive at a contradiction because $a_n = a_n \wedge a = a_n \wedge \bigvee_{i < n} a_i = 0$.
So I am wondering if I am missing an argument that does not depend on completeness of the Boolean algebra. In other words, is the above claim true?
By the way, Johnstone's proof is still fine, because the Boolean algebra there is complete.
Johnstone's claim is true.
Let $B$ be an infinite Boolean algebra. An antichain is a set $\{a_i\mid i\in I\}$ such that $a_i\neq \bot$ for all $i$ and $a_i\land a_j = \bot$ for all $i\neq j$. The argument at the beginning of your proof shows that $B$ admits an infinite antichain. Extend this to an infinite maximal antichain $\{a_i\mid i\in I\}$ by Zorn's lemma.
I claim that $(a_i)_{i\in I}$ has a join, and this join is $\top$. Suppose there is an element $b$ with $b\neq\top$ such that $a_i\leq b$ for all $i\in I$. Then $\lnot b\neq \bot$ and $a_i\land \lnot b = \bot$ for all $i\in I$, so $\{a_i\mid i\in I\}\cup \{b\}$ is an antichain, contradicting maximality. Thus the only upper bound for $(a_i)_{i\in I}$ is $\top$, so $\bigvee_{i\in I} a_i = \top$. Now continue as in your proof to see that there is no finite subset $I_0\subseteq I$ such that $\bigvee_{i\in I_0} a_i = \top$: Suppose there were such an $I_0$, and pick $j\in I\setminus I_0$. Then $$a_j = a_j\land \top = a_j\land \bigvee_{i\in I_0} a_i = \bigvee_{i\in I_0} (a_j\land a_i) = \bot$$ which is a contradiction. So $\top$ is not compact.