It is known that for any $m\in\mathbb{N}$ there exist a set $M_m$ of $m$ points on plane with mutual integer distances. Further it boils down to prove that for any finite set of points $X$ on the plane there exists a point $a$ such that the distance from $a$ to any $x\in X$ is irrational. Indeed, we taka a point $x_0\in X$ and construct a circle $C_0$ with center $x_0$ and radius some irrational number. Take another $x_1\in X$. For each $r\in\mathbb{Q}$ there are at most $2$ points on $C$ at distance $r$ from $x_1$. It means there is uncountable set $C'\subset C$ which points are at irrational distance from $x_1$ (and $x_0$). Proceeding like that we can find uncountable number of points on $C$ each of which is at irrational distance from all the points in $X$. Further, we exploit the fact mentioned. Start from $M_m, m:=n-k$ and choose a point $x_1$ such that the distance from $x_1$ to any point in $M$ is irrational. Set $M':=M\cup\{x_1\}$ and proceed further to finally get pints $x_1,x_2,\dots,x_k$.

To prove the first fact, it suffices to show this for rational distances. Consider the infinitely many points generated by $P_t(\frac{1-6t^2+t^4}{1+2t^2+t^4},\frac{4t-4t^3}{1+2t^2+t^4})$ with $t\in \mathbb{Q}$. It can be checked that all these points have rational distance.

Quote: $P_t(\frac{1-6t^2+t^4}{1+2t^2+t^4},\frac{4t-4t^3}{1+2t^2+t^4})$ My reaction to that information: Different solution