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.

isconstructive (or can easily be made constructive), as pointed out by Valerio Caprano. However, the "obvious" fact that every infinite set contains a countably infinite subset may be seen as nonconstructive by some. And indeed, the theorem you stated is not provable in set theory without the axiom of choice (say: in ZF). (Hint: amorphous sets.) $\endgroup$A Problem Seminar. $\endgroup$