Understanding a detail in a proof related to almost disjoint families (Kunen)

The sets $A$ and $B$ that I constructed in that earlier answer don’t solve the original problem: that question asked how to show Ken’s hint. The hint is this:

$mathcal{A}={a_alpha:alpha<omega_1}$, and $mathcal{B}setminusmathcal{A}={b_alpha:alpha<omega_1}$. Construct $a_alpha,b_alpha$ inductively so that $a_alphacap b_alpha=varnothing$ but $alphanebetato a_alphacap b_betanevarnothing$.

The construction of $A$ and $B$ in that earlier answer is how you construct $a_alpha$ and $b_beta$ when you’ve already constructed $a_xi$ and $b_xi$ for $xi<alpha$. The $mathscr{A}$ in that answer is ${a_xi:xi<alpha}$, temporarily re-indexed by $omega$, and the $mathscr{B}$ is ${b_xi:xi<alpha}$. To finish the proof once you’ve constructed the subsets $a_alpha$ and $b_alpha$ of $omega$, let $$mathcal{A}={a_alpha:alpha<omega_1}$$ and $$mathcal{B}=mathcal{A}cup{b_alpha:alpha<omega_1};.$$ By construction these are almost disjoint families, and $|mathcal{A}|=|mathcal{B}setminusmathcal{A}|=omega_1$, but we still need to show that they satisfy the requirements of the original problem.

Unfortunately, that problem seems to be misstated. In my copy of Ken’s book it reads as follows:

Show (in $mathbf{ZFC}$) that the result of Exercise $9$ can be false if $|mathscr{A}|=|mathscr{B}setminusmathscr{A}|=omega_1$.

(This is then followed by the hint.) The result from Exercise $9$ is that if $mathscr{B}subseteqwp(omega)$ is an almost disjoint family of size $omega$, and $mathscr{A}subseteqmathscr{B}$ is countable, then there is a $dsubseteqomega$ such that $forall xinmathscr{A}(|dcap x|<omega)$ and $forall xinmathscr{B}setminusmathscr{A}(|xsetminus d|color{red}{<}omega)$. Thus, we need to show that there is no $dsubseteqomega$ such that $|dcap a_alpha|<omega$ and $|b_alphasetminus d|<omega$ for each $alpha<omega_1$. (That is, no subset of $omega$ is almost disjoint from each $a_alpha$ and almost contains each $b_alpha$.)

Suppose, on the contrary, that $d$ is such a set. Now $omega$ has only $omega$ finite subsets, so there are finite $r,ssubseteqomega$ and an uncountable $Asubseteqomega_1$ such that $dcap a_alpha=r$ and $b_alphasetminus d=s$ for each $alphain A$. Recall that $a_alphacap b_alpha=varnothing$, so $rcap s=varnothing$. Suppose that $alpha,betain A$, and $alphanebeta$; then $a_alphacap b_betanevarnothing$, so let $nin a_alphacap b_beta$.

If $nin d$, then $nin a_alphacap d=r=a_betacap d$; but then $nin a_betacap b_beta=varnothing$, which is absurd, so $nnotin d$, and therefore $nin b_betasetminus d=s$. And that is also impossible: then $nin s=b_alphasetminus d$ and hence $nin a_alphacap b_alpha=varnothing$. Thus, there is indeed no such $dsubseteqomega$.