TITLE: Uncountable family of infinite subsets with pairwise finite intersections
QUESTION [8 upvotes]: 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.

REPLY [12 votes]: Here is a transcription of Todd Eisworth's answer into constructive mathematics. His construction is more or less constructive, he just uneccessarily says things which ruin the constructivity such as:

In constructive mathematics only decidable subsets of $\mathbb{N}$ have characteristic functions. But we can avoid the problem by speaking only about the characteristic functions.
Not every countable set can be put in bijective correspondence with the natural numbers. This can be avoided, because we don't need such a bijection, just an injection from the natural numbers into the set (which is the definition of "infinite").
In constructive mathematics cardinals don't work too well, so it is better not to mention $2^{\aleph_0}$. This can be avoidded by talking explicitly about a set being uncountable (i.e., there is no surjection from $\mathbb{N}$ onto the set).

And here is the proof:
The set $2^\ast$ of finite sequences of $0$'s and $1$'s is in bijective correspondence with $\mathbb{N}$, therefore it clearly suffices to find an uncountable collection of subsets of $2^\ast$ such that any two of them have only a finite intersection. Once such a collection is found, it can be embeded into $X$.
The set $2^\mathbb{N}$ is uncountable by the usual Cantor's diagonal argument (which is constructive!). Given any $f \in 2^\mathbb{N}$, let $S_f \subseteq 2^\ast$ be the set of finite prefixes of $f$. The assignment $f \mapsto S_f$ is injective, therefore the $S_f$'s form an uncountable family. If $f$ and $g$ are distinct, then there is a smallest $n \in \mathbb{N}$ such that $f(n) \neq g(n)$, hence $S_f \cap S_g$ is finite, as it contains precisely the first $n$ prefixes of $f$.

REPLY [3 votes]: There are several constructions, some of which are quite nice and visual, and which you are therefore more likely to consider "constructive". These examples were shown to me by Imre Leader.

Your example: take your infinite set to be $\mathbf{Q}$ and look at approximations to reals.
Take your infinite set to be the nodes of an infinite binary tree, and take your sets to be the paths from the root to infinity. Every two such paths will go separate ways after finitely many steps.
Take your infinite set to be the quarter-lattice $\mathbf{N}^2$, and take your sets, indexed by $\theta\in[0,\pi/2)$, to consist of the lattice points between $y=x \tan(\theta)$ and $y=x\tan(\theta) + 1$. Any two such slivers have finite intersection.

Complementary question: Find an uncountable totally ordered chain of subsets of $\mathbf{N}$.