Prove that for each sufficiently large integer $N$ there do not exist positive integers $a,q$ with $a<q, (a,q)=1$ such that $q\leq N^{\frac{1}{20}}$ and the distance from $\sqrt{2} - \frac{a}{q}$ to its nearest integer is at most $N^{-\frac{9}{10}}$.
(NOTE: In the original post the expression "do not" was missing, now it's fixed and a solution in given in one of the answer posts. Gratitude to John Don for this.)
Under a specific definition of minor/major arcs in Waring's problem for sums of squares, this is equivalent to $\sqrt{2}$ lying in a major arc. I am definitely not aiming for a solution with involved theory around this, I just mentioned where the problem comes from.
I am given as a hint to use that $|a^2 - 2q^2|\geq 1$ if $a,q$ are not both zero but apart from transforming into $|\sqrt{2}-\frac{a}{q}||\sqrt{2}+\frac{a}{q}| \geq 1$, I do not know how to continue.
Any help appreciated!
There is actually a mistake in the question - $\sqrt{2}$ does lie in a minor arc (and in fact your proof was headed in the right direction!).
As you note, for all $a, q$ non-zero, $|a^2 - 2q^2| \geqslant 1$, and upon dividing through by $q^2$ and taking the difference of two squares, we get $|\sqrt{2} - \frac{a}q||\sqrt{2} + \frac{a}q| \geqslant \frac1{q^2}$. $\,\,$($\dagger$)
Now if $\sqrt{2}$ were in a major arc, then we would have that $|\sqrt{2} - \frac{a}q| \leqslant N^{-\frac9{10}}$ for some $q \leqslant N^{\frac1{20}}$ and $(a, q) = 1$ (this is just a slight rephrasing of the condition as given in your question). Note that, for such integers $a$ and $q$, $|\sqrt{2} + \frac{a}q| < 4$.
But this would give us $|\sqrt{2} - \frac{a}q||\sqrt{2} + \frac{a}q| < 4N^{-\frac9{10}} \,$ which is less that $\frac1{q^2}$ for all $q \leqslant N^{\frac1{20}}$ (if $N$ is sufficiently large), contradicting ($\dagger$).