I recently stumbled upon the concept of almost disjoint families and I read that maximal almost disjointness requires at least an uncountable cardinality of the family. I believe it is not so clear to me why this is the case. Here is my try:
Let $\mathcal{A}$ be a countable almost disjoint family. Assume that $\mathcal{A}$ is infinite. Then $\mathcal{A} = \{ A_1, A_2, \dots \}$ with $A_1 = \{ a_{11}, a_{12}, \dots \}$, $A_2 = \{ a_{21}, a_{22}, \dots \}$, and so on. From the $A_k$ construct a new set in $\mathcal{A}$. To do so one has to make sure that this new set $B$ is different from each of the $A_k$ and has finite intersection with each of them. Take $B=\{ b_1, b_2, \dots \}$, $b_k \in A_k \setminus \bigcup_{l=1}^{k-1}A_l$.
Surely there's details missing in the above try, but is it in principle ok?
If $\mathcal{A}$ is finite, then the above does not seem to work. For example, pick $\mathcal{A}$ to be the partition of two subsets of $\mathbb{N}$, the even and the odd numbers. But then I cannot find an almost disjoint family which strictly contains $\mathcal{A}$, because by the pigeonhole principle, any element of an infinite subset $B \subseteq \mathbb{N}$ intersects at least one of the two sets in the partition infinitely often. What am I missing?