Proving there are infinitely many rational numbers in $[x,y]$ 
Prove if $x$ and $y$ are real numbers with $x \lt y$, then there are infinitely many rational numbers in the interval $[x,y]$.

What I got so far:
Let $x,y \in \Bbb R$ with $x \lt y$ 
Let $S = [x,y]$
By the density of $\Bbb Q$ in $\Bbb R$, $\exists r \in \Bbb Q$ such that $x \lt r \lt y$ where $r \in S$.

This is where I got stuck. 
 A: If you have only $n$ finitely many rational numbers between $x$ and $y$, put them in order, $x, r_1, r_2, \ldots r_n, y$.  Can you now prove that there were more than $n$ rational numbers between $x$ and $y$?  If so, you have contradicted your original assumption that there were only $n$ such numbers, which means no such $n$ can exists and there must be infinitely many rational numbers between $x$ and $y$.
A: Just find two distinct rational numbers $p,q\in[x,y]$. Now the rational number $p+(q-p)/n$ is in $[x,y]$ for all $n\in \Bbb N_{>0}$.
(And you can find such $p,q$ in the intervals $[x,x+(y-x)/3]$ and $[x+2(y-x)/3,y]$.)
A: Okay, so I'll give it another shot given the feedback.
Proof:
Let $x,y \in \Bbb R$ with $x \lt y$ and $S = [x,y]$
Suppose there are only $n$ rational numbers between $x$ and $y$ such that:$$x \lt r_1 \lt \cdot \cdot \cdot \lt r_n \lt y$$
But since $\Bbb Q$ is dense in $\Bbb R$, there exists $r_{n+1} \in \Bbb Q$ such that: $$ x \lt r_{n+1} \lt r_1  \lt \cdot \cdot \cdot \lt r_n \lt y$$
which contradicts our assumption that there are only $n$ rational numbers in $[x,y]$.
Therefore, there must be infinitely many rational numbers in the interval $[x,y]$.
A: Consider the interval $[x,y]$.  Find two rational numbers $q_{min}$ and $q_{max}$ such that $x < q_{min} < q_{max} < y$.  Let $r=q_{min}-q_{max}$.  Now make a new number: $q_1 = q_{min} + r/2$.  Let $r_1=q_{max}-q_1$.  Now make a new number: $q_2=q_1+r_1/2$.  Let $r_2=q_{max}-q_2$.  Follow the pattern.  Each of these numbers is a rational number; each number is unique.  The set constructed is countably infinite in size.
