Premise of Minkowski's theorem: "Irrational" Line If we take a straight line $L$ in $\Bbb{R}^2$ of the form $y=ax$ where $a\in\Bbb{R}\setminus{\Bbb{Q}}.$

Then we see that (except at the origin $(0,0)$) $$L\cap \Bbb{Z}^2=\emptyset.$$

But now if the line becomes a strip of width $r&gt;0$ denoted $S$ how can we be sure that $$S\cap \Bbb{Z}^2\ne\emptyset?$$

I know about Minkowski's theorem but in this particular case is there a high school argument? Perhaps by writing the continued fraction of $a?$
Let $m\in\mathbb{N}$ be such that $1/m&lt;r$.  Given $m+1$ real numbers, there must be two of them whose fractional parts are within an interval of length $1/m$, by pigeonhole.  In particular, there must be two different integers $n$ and $n'$ such that the fractional parts of $an$ and $an'$ differ by less than $1/m$.  Then $a(n-n')$ will be within $1/m$ of an integer $k$.  The point $(n-n',k)\in\mathbb{Z}^2$ is then in your strip $S$, since it is within $r$ of $(n-n',a(n-n'))$.
The strip is an open set so it contains a rational point $q = (a/b, c/d)$, because rationals points are dense in Euclidean space. Since it is a line, and in particular closed under scaling, it contains $(bd)q = (a,c)$ which is an integer point.

Edit: obviously this doesn't work, since it's not closed under scaling. I was being sloppy. Thanks to mercio for pointing that out. I'll leave this here to embarrass me and encourage me to check my work.