Rational contained in open interval centered at an irrational. 
Lemma. Let $y \in [0,1] \setminus \mathbb{Q}$,  $\epsilon > 0$ and $N \in \mathbb{N}$ such that $\frac{1}{N} < \epsilon$.
Then one can choose $\delta > 0$ small enough such that
$$  \frac{p}{q} \in (y - \delta, y + \delta) \Rightarrow q > N, $$
where $\frac{p}{q} \in \mathbb{Q}$ and $gcd(p, q) = 1$.

I can see that any $\frac{p}{q}$ that satisfies the condition equals $y + t$, where $\delta \le |t| \in [0,1] \setminus \mathbb{Q}$. I just don't see how we can make $\frac{1}{q}$ arbitrarily small.
 A: I'll outline how to prove the larger theorem:

If $y \in [0,1] \setminus \mathbb{Q},\quad p,q\in\mathbb{N},\ q\neq 0\ $ and $\ N\in\mathbb{N}\ $ then
one can find $\ \delta>0\ $ small enough so that
$$  \frac{p}{q} \in (y - \delta, y + \delta) \Rightarrow q > N.$$

This encompasses your theorem, as the $\ gcd\ $ condition is satisfied also.

*

*Use the fact that $\ y\ $ is irrational to show:

$\exists \delta_1>0\ $ such that $\ \frac{1}{1},\ \frac{2}{1},\ \frac{3}{1},\ \ldots\ $ are all outside $\ (y-\delta_1, y+\delta_1)$;
$\exists \delta_2>0\ $ such that $\ \frac{1}{2},\ \frac{2}{2},\ \frac{3}{2},\ \ldots\ $ are all outside $\ (y-\delta_2, y+\delta_2)$;
$\exists \delta_3>0\ $ such that $\ \frac{1}{3},\ \frac{2}{3},\ \frac{3}{3},\ \ldots\ $ are all outside $\ (y-\delta_3, y+\delta_3)$;
$\ldots$
$\exists \delta_{N}>0\ $ such that $\ \frac{1}{N},\ \frac{2}{N},\ \frac{3}{N},\ \ldots\ $ are all outside $\ (y-\delta_{N}, y+\delta_{N}).$

*

*Let $\ \delta = \min\{\delta_1,\delta_2,\ldots,\delta_{N+1}\}.$


*If $\ q\in\mathbb{N}\ $ and $\ q<N+1,\ $ then $\ \frac{p}{q}\ \notin (y-\delta, y+\delta).$


*The contrapositive statement, that is, $\ \frac{p}{q}\ \in (y-\delta, y+\delta) \implies q \geq N+1\ $ must also be true.
