I'm working through this proof and I'm having trouble justifying this step because the book briefly goes over it. I know you have to use the regularity of $\kappa$ but I don't know how to set it up.
Let $\kappa$ be an uncountable regular cardinal and let $\mathcal{A}$ be a family of finite sets with $|\mathcal{A}| = \kappa$. Since $\text{cf}(\kappa) = \kappa > \omega$ and there are only $\aleph_0$ possible $|X|$ for $X\in \mathcal{A}$, we may fix $n\in \omega$ and $\mathcal{D}\in [\mathcal{A}]^\kappa$ such that $|X| = n$ for all $X\in \mathcal{D}$.
In other words, given a family of $\kappa$ finite sets, there exists a subset of the same cardinality in which every set in the subset has exactly $n$ elements.
How do you use the regularity of $\kappa$ to show this? What is the cofinal sequence that we use here to show the result?
We show this by contradiction. Let $\mathcal{A}$ be a family of finite sets such that $|\mathcal{A}| = \kappa$ for some uncountable regular $\kappa$. Define $\mathcal{D}_n$ as the family of all $X\in \mathcal{A}$ with exactly $n$ elements. Suppose every $|\mathcal{D}_n|< \kappa$. Note that $|\bigcup \mathcal{D_n}| = |\mathcal{A}| = \kappa$. But there are at most $\aleph_0$ many $\mathcal{D}_n$. Because $\kappa$ is regular, $|D_n| < \kappa$ and $|\{\mathcal{D}_n : n\in \omega\}| <\kappa$ implies $|\bigcup\mathcal{D_n}|< \kappa$. But this is a contradiction. Thus, there must be at least one $\mathcal{D_n}$ with cardinality $\kappa$.