Here is a theorem:
Let $E$ be an uncountable infinite set. Then there is a collection $\mathcal{A}$ of subsets of $E$ such that $|\mathcal{A}|=|E|^\omega$, $|A|=\omega$ for each $A\in \mathcal{A}$, and the intersection of any two distinct elements of $\mathcal{A}$ is finite.
How to show it? I hope the proof a little simple and clear, because it is too difficult in my textbook. Thanks.
ADD: Thanks Brain. In my textbook, it said, it suffices to construct the desired collection on the set of all finite subset of $\omega \times E$. For each $f\in {}^\omega E$, let $A(f)=\{(f|n):n<\omega\}$. One can easily check $A(f)\cap A(g)$ is finite whenever $f$ and $g$ are distinct elements of ${}^\omega E$, so $\{A(f): f\in {}^\omega E\}$ is the desired collection of sets. I cannot check this sentence: One can easily check $A(f)\cap A(g)$ is finite whenever $f$ and $g$ are distinct elements of ${}^\omega E$.
Let $T={}^{<\omega}E=\bigcup_{n\in\omega}{}^nE$, the set of functions from finite ordinals into $E$. Then $\langle T,\subseteq\rangle$ is a tree of height $\omega$. Note that for $s\in{}^mE$ and $t\in{}^nE$, $s\subseteq t$ iff $t\upharpoonright m=s$. Each $\sigma\in{}^\omega E$ defines a branch through $T$; for $\sigma\in{}^\omega E$ let $B_\sigma=\{\sigma\upharpoonright n:n\in\omega\}$, the set of nodes on that branch. Each $B_\sigma$ is a countably infinite subset of $T$. $|T|=\omega\cdot|E|=|E|$, so there is a bijection $\varphi:T\to E$; for each $\sigma\in{}^\omega E$ let $A_\sigma=\varphi[B_\sigma]$, and let $\mathscr{A}=\left\{A_\sigma:\sigma\in{}^\omega E\right\}$. I leave it to you to verify that $\mathscr{A}$ has the desired properties.