Let $E/k$ be an elliptic curve defined by the Weierstrass form $y^2=x^3+ax+b$. Let $c$ be a nonzero squarefree element in $k$. Let $E_c/k$ be a curve defined by $cy^2=x^3+ax+b$. Show that the 2-torsion subgroup $E[2](k)=E[2]\cap E(k)$ of $E(k)$ is the same as the 2-torsion subgroup $E_c[2](k)=E_c[2]\cap E_c(k)$ of $E_c(k)$.
I know and have shown that $E_c$ is a twist of $E$ but I'm not sure how to use that here. What's an easy way to show these 2-torsion subgroups are equivalent?
For both curves, the $2$-torsion points are the points $(x_0,0)$ such that $x_0^3+ax_0+b=0$ (together with the point at infinity of course).