I faced following problem:
Prove that there is an s.d. family $\mathscr{A}\subset\mathcal{P}(\omega)$ of size $\omega_1$ which has a subfamily $\mathscr{B}\subseteq\mathscr{A}$ s.t. $|\mathscr{B}|=|\mathscr{A-B}|=\omega_1$. If $|d\cap x|<\omega$ for each $x\in\mathscr{B}$ then there is $y\in \mathscr{A-B}$ s.t. $|y-d|=\omega$.
Hint. Construct $a_\alpha$, $b_\alpha$ inductively so that $a_\alpha\cap b_\alpha=\varnothing$ but $a_\alpha\cap b_\beta\neq\varnothing$ if $\alpha\neq\beta$.
I construct $a_\alpha$, $b_\alpha$ which satisfy the properties given from the hint. But I don't know how to prove the problem using $a_\alpha$ and $b_\alpha$. I would appreciate any help.
(I modify the question.)
I think the problem should say: Construct $\mathcal{B} \subseteq \mathcal{A}$, such that ...
The hint is saying that letting $\mathcal{B} = \{b_{\alpha} : \alpha < \omega_1\}$, $\mathcal{A} - \mathcal{B} = \{a_{\alpha} : \alpha < \omega_1\}$ should work where $\mathcal{A}$ is a.d. family and for all $\alpha \neq \beta$, $a_{\alpha} \cap b_{\alpha} = \phi$, $a_{\alpha} \cap b_{\beta} \neq \phi$. To see this suppose $d$ is infinite and almost disjoint with each $b_{\alpha}$. By deleting finitely many elements from $d$, we can assume that it is disjoint with $b_{\alpha}$ for all $\alpha \in X$ where $X$ is an uncountable subset of $\omega_1$. Suppose $|a_{\alpha} - d| < \omega$ for every $\alpha$. Then for some finite set $F$ and an uncountable $Y \subseteq X$, we have $a_{\alpha} \subseteq d \cup F$ for every $\alpha \in Y$. Now choose $\alpha < \beta \in Y$ such that $a_{\alpha} \cap F = a_{\beta} \cap F$. Then, $a_{\alpha} \cap b_{\beta} = a_{\alpha} \cap b_{\beta} \cap F = a_{\beta} \cap b_{\beta} = \phi$ which is impossible.