I was able to prove the lemma and now I am proving a). I am confused for the highlighted red part why $\displaystyle \cup_{A \in \mathcal{A}} J_A$ would be finite. I thought maybe I should see if there's a contradiction if $\displaystyle \cup_{A \in \mathcal{A}} J_A$ was infinite. Since there are finitely many elements in $\mathcal{A}$, it must be that for some $B \in \mathcal{A}$, $J_B$ is infinite. This would mean that for infinitely many indices $i < \beta$, $\pi_i(B) \neq X_i$. But I don't see anything bad happening if that was the case.
All sets $J_A$ are by definition finite (this is what $A$ being basic open means). As $\mathcal{A}$ is finite, the union is clearly finite.