Paper Punches puzzle!

Note that punching the plane at a center will leave only the points that lie on the circles concentric with center the punching point and radii the rational numbers.

So clearly, $2$ punches is not sufficient since the intersection of the two sets of concentric circles still leaves points that are not removed. But is it possible to remove all points with three punches?

I claim that punching the plane at the origin and $(\pm \pi,0)$ will remove all points.

Suppose that it doesn't. Then a point $P(x,y)$ satisfies that the distance between $P$ and the three punch centers are all rational.

So $\sqrt{x^2+y^2}=\dfrac{p_1}{q_1}$ $\sqrt{(x-\pi)^2+y^2}=\dfrac{p_2}{q_2}$ $\sqrt{(x+\pi)^2+y^2}=\dfrac{p_3}{q_3}$ for integers $p_k, q_k$ .

Squaring and adding the last two equations gives $x^2+y^2 +2\pi^2= \dfrac{p_2^2}{q_2^2}+\dfrac{p_3^2}{q_3^2}$

Squaring the first equation gives $x^2+y^2=\dfrac{p_1^2}{p_2^2}$

Substituting $x^2+y^2$ and simplifying gives $2\pi^2=\dfrac{p_2^2}{q_2^2}+\dfrac{p_3^2}{q_3^2}-\dfrac{p_1^2}{p_2^2}$

But $RHS\in\mathbb{Q}$ and $LHS\not\in\mathbb{Q}$ contradiction.

Thus, $\boxed{3}$ punches are necessary and sufficient.