Prove that a subset of an infinite subset of $\mathbb{R}$ is countable. Let $A$ be an infinite subset of $\mathbb{R}$. Find a subset $H$ of $A$, and prove that it is countable.
Since I do not know what $A$ is, to form a subset of it, I am going to define a set $H$ that is comprised of $a_{q,n}$, which are elements of the intersection of $A$ and $I_{q,n}$, where for each $q \in \mathbb{Q}$ and each $n \in \mathbb{N}$, $I_{q,n} = (q-\frac{1}{n}, q+\frac{1}{n})$. If the intersection turns up empty, let $a_{q,n}$ go undefined. So, in other words, $H=\{ a_{q,n} : q \in \mathbb{Q}, n \in \mathbb{N} \}$.
$H$ is certainly a subset of $A$, but how do I show that $H$ is countable? I think I need to find an injective function $f:H -> \mathbb{Q} cross \mathbb{N} $, but I am not sure how to do that.

SKETCH: As Asaf said, the fact that you’re working in $\Bbb R$ is irrelevant; just let $A$ be any infinite set. Construct $H$ recursively, one point at a time. The fact that $A$ is infinite means that at every stage, when you’ve picked only the first $n$ points, say, there’s sure to be a point left to choose. Try to write it up yourself. If you get stuck, I’ve left a spoiler-protected write-up below; mouse-over to see it.

 Let $a_0\in A$ be arbitrary. Since $A$ is infinite, $A\setminus\{a_0\}\ne\varnothing$, so we may choose any $a_1\in A\setminus\{a_0\}$. In general suppose that $n\in\Bbb N$, and we have already chosen distinct elements $a_k\in A$ for $k<n$; $A$ is infinite, so $A\setminus\{a_k:k<n\}\ne\varnothing$, and we may choose $a_n\in A\setminus\{a_k:k<n\}$. In the end let $H=\{a_n:n\in\Bbb N\}$. By construction the map $\Bbb N\to H:n\mapsto a_n$ is a bijection.

