$B$: the complete Boolean algebra. $V^B$: the $B$-valued model.
Lemma 14.18. If $W$ is a set of pairwise disjoint elements of $B$ and if $a_u$ (where $u\in W$) are elements of $V^B$, then there exists some $a\in V^B$ such that $u\leq||a=a_u||$ for all $u\in W$.
Lemma 14.19 $V^B$ is full. Given a formula $\varphi(x,\ldots)$, there exists some $a\in V^B$ such that $||\varphi(a,\ldots)||=||\exists x\varphi(x,\ldots)||$.
Proof. We wish to find an $a\in V^B$ such that $\geq$ holds Let $u_0=||\exists x\varphi(x,\ldots)||$. Let $$ D=\{u\in B:\textrm{there is some }a_u\textrm{ such that }u\leq||\varphi(a_u,\ldots)||\}. $$ It is clear that $D$ is open and dense below $u_0$. Let $W$ be a maximal set of pairwise disjoint elements of $D$; clearly, $\sum\{u:u\in W\}\geq u_0$. By Lemma 14.18 thre exists some $a\in V^B$ such that $u\leq\||a=a_u||$ for some $u\in W$. Thus for each $u\in W$ we have $u\leq||\varphi(a,\ldots)||$, and hence $u_0\leq||\varphi(a,\ldots)||$.
I don't know why we need $D$ is open. More specifically, if $D$ is dense below $u_0$, I think we would have the followings:
(i) $\sum D=u_0$
(ii) $\sum W$= $\sum D$.
Then $\sum W=u_0$ and the left proceeding still works.
You don't need that $D$ is open and are correct about all three statements.
I can only guess why Jech mentions that $D$ is also open here but, in any case, you'll find this sort of unnecessary information in mathematical proofs all the time for a plethora of possible reasons.