Infinite number of rationals between any two reals.

Let $a$ and $b$ be reals with $a<b$. Show that there are infinitely many rationals $x$ such that $a<x<b$.

My plan of action was to assume that $x$ is the smallest such rational and find another rational in the interval $(a, x)$, but I am struggling to make it work. A hint will be much preferred to a full solution.

Note that the real numbers are an Archimedean field, so for any real number $r$ we have some integer $n>r$. This means that for any real number $\epsilon>0$, we have some $n>1/\epsilon$, so $1/n<\epsilon$. Furthermore, the rationals are dense in the reals, so we can find some rational $x$ such that $a<x<b$.

Let $n$ be such that $1/n<b-x$. Then $x+\frac{1}{n},x+\frac{1}{n+1},\ldots$ is an infinite set of rationals between $a$ and $b$.

Maybe one can try something like this. Using the following

Lemma 1. For every real number $x$ there is exactly one integer $N$ such that $N \le x < N + 1$. (This integer $N$ is called the integer part of $x$, and is sometimes denoted $N = \lfloor x \rfloor$.)

Lemma 2. For any positive real number $x > 0$ there exists a positive integer $N$ such that $0 < 1/N < x$.

We now show

Proposition 3. Given any two real numbers $x < y$, we can find a rational number $q$ such that $x < q < y$.

By hypothesis, we have $y -x$ is positive. By Lemma 2, exists a positive integer $N$ such that $0 < 1/N < y - x$. Since $xN$ is a real number, by Lemma 1, there exists a integer $n$ such that $n - 1 \le xN < n$, i.e., $n/N - 1/N \le x$ and $x < n/N$. Thus $x < n/N \le x + 1/N$. Since $1/N \le y - x$, i.e., $x + 1/N < y$, we have $x < n/N < y$. Thus $n/N$ is rational, the claim follows.