A classical one:
Problem. There exists an uncountable family $\{X_i: i \in I\}$ of subsets of $\mathbb{N}$ such that each $X_i$ is infinite and $X_i \cap X_j$ is finite for all distinct $i,j \in I$.
Proof. Let $f: \mathbb{Q}\cap [0,1] \to \mathbb{N}$ be a bijection. Fix each $r \in [0,1]$, let $(x_{r,n}: n\ge 0)$ be a sequence of rationals in $[0,1]$ which is convergent to $r$ with $r\neq x_{r,n}$ for all $n$. Then the family $\{f[\{x_{r,n}: n\ge 0\}]: r \in [0,1]\}$ satisfies the claim.
Now the two-dimensional analogue:
Question. Does there exist an uncountable family $\{X_i: i \in I\}$ of subsets of $\mathbb{N}^2$ such that $X_i \cap (\{k\}\times \mathbb{N})$ is infinite for all $i \in I$ and $k \in \mathbb{N}$, and $X_i \cap X_j$ is finite for all distinct $i,j \in I$?
If $\{X_i : i \in I\}$ is an uncountable almost disjoint family on $\mathbb{N}$, then consider $\{ \bigcup_{n \in\mathbb{N} } \{n\} \times (X_i \setminus n) : i \in I \}$.