On page 9 of "Principles of Mathematical Analysis" by Walter Rudin it is stated that if $x\in\mathbb{R}$, $y\in\mathbb{R}$ and $x<y$ then there exists a $p\in\mathbb{Q}$ such that $x<p<y$ then $\mathbb{Q}$ is dense in $\mathbb{R}$.
How would one prove that obeying this condition implies density by the following definition of density?
For a normed ring $(R,N)$ and $X\subseteq R$, $X$ is said to be dense in $R$ if every element of $R$ is a limit with respect to $N$ of a sequence of elements of $X$.
Here he refers to the usual absolute value norm given by~ $|x|:={\begin{cases}x,&{\text{if }}x\geq 0\\-x,&{\text{if }}x<0.\end{cases}}$
Any help appreciated.
For any $x$ there is rational number $r_n$ in $(x-\frac 1n,x+\frac 1n)$. Check that $|r_n- x| \to 0$.