I am working through my books notes trying to learn on my own and came across the following theorem where the proof is left ot the reader.
Theorem: A rational number is approximated to order 1 and to no higher order.
Definition: A number $\xi $ is aporximable by rationals to order n if there exists a real number $K(\xi) $ s.t $$| \xi-\frac {p}{q}| < \frac {K(\xi)}{q^n }$$ has infinitely many solutions $\frac {p}{q} $
Edit: we have that $\xi \geq 0 $ then if we select $\frac {p}{q} >0 $ then $| \xi-\frac {p}{q}| < \frac {p}{q} $ so select $K(\xi)=2 $ then $| \xi-\frac {p}{q}| < \frac {p}{q} \leq \frac {2}{q}$ where p =1 or 2 and $ q\in \Bbb N $ this gives infinite solutions for the $n=1$ case.
Now let us select $\xi = \frac {a}{b} \geq 0$ notice that $\frac {K(\xi)}{q^2 } >\frac {K(\xi)}{q^3 } > ... $ so we need only show the case $n=2$ cannot have infinite solutions. we have that $ |\frac {a}{b} - \frac {p}{q} |= |\frac {aq- pb }{bq}| \geq \frac {1}{bq}$ now for any choice $ K(\xi ) $ we deifne $q^*=b\lceil K(\xi) \rceil $ the we have that $\frac {1}{bq}= \frac{1}{b^2 \lceil K(\xi) \rceil} <\frac {K(\xi)}{q^2 } < \frac {1}{b^2\lceil K(\xi) \rceil}$ which is a contradiction but there are only finitely many q smaller than $ b \lceil K(\xi) \rceil $ and for those q there is only finitely many p the equation could hold for let $p= (\lceil K(\xi) \rceil q +\lceil \frac{a}{b} \rceil q) $ is an upper bound which is finite so there is only finite p positive less than that.
$K(\xi)$ depends only on $\xi$ and $n$. Your error is that the choice you want to make for $K(\xi)$ depends on $p$ as well.