I'm trying to show that for every real number $r$, there exists a sequence of rational numbers $\{q_{n}\}$ such that $q_{n} \rightarrow r$.
Could I get some comments on my proof?
I know that between 2 reals $r, b$ there exists a rational number $m$ such that $r < m < b$. So I can write
$r < q_{1} < b$ ; Now check if $q_{1} = r$ or not. If it does I'm done, and if not, I consider the interval $(a, q_{1})$.
$r < q_{2} < q_{1}$ check if $q_{2} = r$ or not. If it does I'm done, and if not, I consider the interval $(a, q_{3})$
If I continue in this manner, I see that $|r - q_{n+1}| < |r - q_{n}|$. So whether $r$ is rational or irrational, I'm making my the size of the interval $(r, q_{n})$ closer to 0 and as $n \rightarrow \infty$. And so given any $\epsilon > 0$, I know that $|q_{n} - r| < \epsilon$.
Revision
Let $\{q_{n} \}$ be a sequence of rational numbers and $q_{n} \rightarrow r$ where $r \in \mathbb{R}$.
If $r$ is rational, then let every element of $q_{n} = r$.
But if $r$ is irrational, then consider the interval $(r, b)$ where $b \in \mathbb{R}$. Since we can always find a rational number between two reals, consider
$r < q_{1} < b$. Now pick $q_{2}$ such that $q_{2}$ is the midpoint of $r$ and $q_{1}$. So we get that $r < q_{2} < q_{1}$. Then repeat the process so that $r < q_{n} < q_{n-1}$. Note that $|r - q_{n}| = \frac{1}{2}|r - q_{n-1}|$. As we take more values for $q_{n}$, it is clear that $|q_{n} - r| \rightarrow 0$. So given any $\epsilon > 0$, $|q_{n} - r| < \epsilon$.