Let $\alpha = 0.10110111011110....$ be given real number written in base $10$, i.e., $n-$th digit of $\alpha$ is $1$, unless $n$ is of the form $\dfrac{k(k+1)}{2}$ in which case it is $0$. Then how to prove the following statement?
For every integer $q \geq 2$, there exist an integer $r \geq 1$ such that $$\frac{r}{q} < \alpha < \frac{r+1}{q}$$
I know $\alpha$ is an irrational number. But don't know how to prove this statement?
It follows (but see the note below) immediately from $\alpha$ being irrational:
Suppose $\theta$ is positive and irrational and $q\ge 2$ is an integer. Let $r$ be the largest nonnegative integer such that ${r\over q}\le\theta$ (why does this exist?). Then ${r+1\over q}>\alpha$ (why?); but also, ${r\over q}<\theta$ since $\theta$ is irrational, hence not equal to ${r\over q}$.
So for every positive irrational $\theta$ and integer $q\ge 2$, there is some (unique, obviously) nonnegative integer $r$ such that ${r\over q}<\theta<{r+1\over q}$.
Now in your case you want $r\ge 1$; so it's enough to show that ${1\over q}<\alpha$ for all $q\ge 2$. Do you see how to do this?
Note: at this point, an error becomes clear in the question - ${1\over 2}$ is not $<\alpha$ as defined. Presumably $\alpha$ is in binary, as opposed to decimal? Alternatively, did you just want $r\ge 0$ instead of $r\ge 1$?