I am searching for a constructive proof of the following fact: If $X$ is an infinite set, there exists an uncountable family $(X\_\alpha)\_{\alpha \in A}$ of infinite subsets of $X$ such that $X\_\alpha \cap X\_\beta$ is finite whenever $\alpha \neq \beta$. The way I know how to prove this statement is as follows.


First, it suffices to prove the case when $X$ is countable. Thus we can choose a bijection between $X$ and $\mathbb{Q} \cap [0,1]$. To save notation we can tacitly assume that $X = \mathbb{Q} \cap [0,1]$. 


Let the index set be $A = [0,1] \setminus X$, i.e. all the irrationals in $[0,1]$. For each $\alpha \in A$, choose a sequence $(x\_{\alpha 1},x\_{\alpha 2},\dots)$ of elements of $X$ such that $x\_{\alpha n} \to \alpha$ as $n \to \infty$, and let $X\_\alpha = \{ x\_{\alpha\_n} \mid n \in \mathbb{N} \}$.


Since $\alpha$ is irrational, the sequence $(x\_{\alpha n})$ cannot be eventually constant, so $X\_\alpha$ is infinite. And if $\alpha \neq \beta$ then the sequences $(x\_{\alpha n})$ and $(x\_{\beta n})$ can have only finitely many terms in common since they have different limits, so $X\_\alpha \cap X\_\beta$ is finite.


Is it possible to do this in a more constructive way? I know very little about set theory and logic, so I apologize if this question is too elementary. Also, I wasn't sure about any relevant tags other than set-theory, so please feel free to add appropriate tags.


Edit: to clarify, I didn't have a clear notion of what I meant by "constructive" here. What I didn't like about the proof I gave above was that it required a choice of sequence of rationals converging to each irrational. The answers so far all address this concern adequately.



Uncountable family of infinite subsets with pairwise finite intersections
This all hinges on what you mean by constructive. An easy way to get such a family is to proceed as follows:


Put your countable set $X$ in bijective correspondence with the collection of finite sequences of 0s and 1s.


For every every subset $A$ of the natural numbers, let $\chi\_A:\mathbb{N}\rightarrow\{0,1\}$ be the characteristic function of $A$, and let $S\_A$ be the collection of finite sequences of the form $\chi\_A|\{0,\dots,n\}$ for $n\in\mathbb{N}$. 


Each $S\_A$ is infinite, and if $A$ and $B$ are distinct subsets of $\mathbb{N}$, then $S\_A\cap S\_B$ is finite (once we get past the first difference between $A$ and $B$, the characteristic functions disagree). Now use your bijection to pull things back inside your original set $X$. This gives you a family of size $2^{\aleph\_0}$ enjoying the property you want.


I don't know if this is constructive enough or not!


