Pick the Cartesian point $ P=\left( 1/7, \sqrt 3 \right) . $ Prove that no two lattice points are equidistant from $ P. $ The circle with center $ P $ and radius $ 2 $ contains exactly one lattice point. Now, suppose there is a circle centered at $ P $ containing $ n $ points. Increase its radius until it touches another lattice point (this happens because adding a number between $ 0 $ and $ \sqrt 2, $ the circle passes the vertex of an unit square which has diagonal $ \sqrt 2 $). Having in mind that no two points are equidistant from $ P, $ we´ve obtained a circle containing exactly $ n+1 $ points. By induction, we're done.

I think, it's meant $n$ lattice point on the circumference of the circle, not inside the disk. Otherwise, it's straightforward, as shown above. So, I think they proposed it having in mind Schinzel's theorem. Btw, the existence of a point $P$, no two lattice points are equidistant from $P$, could be shown non-constructively in the following way. Consider the family $F$ of all possible perpendicular bisectors of $XY$ where $X,Y$ are lattice points. Clearly these lines are countably many. We claim they cannot cover all points of the plane. Indeed, consider some circle $C$. Any line from $F$ intersects $C$ in at most $2$ points, thus the set of points the lines from $F$ intersect $C$ is countable. But all points of $C$ are uncountably many, so there exists $P\in C$ that do not belong to any line in $F$.