jmc

To find the $x$-coordinates of the intersections, substitute $x^2$ for $y$ in $x+y=1$ and solve for $x$, resulting in $$\begin{array}{rl}x+x^2& =1\\ \Rightarrow x^2+x-1& =0\\ \Rightarrow x& =\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}\end{array}$$Using each of these coordinates to solve for $y$ gives us the intersections at $\left(\frac{-1+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}\right)$ and $\left(\frac{-1-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}\right)$. Using the distance formula, we have $$\begin{array}{rl}& \sqrt{{\left(\frac{-1+\sqrt{5}}{2}-\frac{-1-\sqrt{5}}{2}\right)}^2+{\left(\frac{3-\sqrt{5}}{2}-\frac{3+\sqrt{5}}{2}\right)}^2}\\ & =\sqrt{{\left(\frac{2\sqrt{5}}{2}\right)}^2+{\left(-\frac{2\sqrt{5}}{2}\right)}^2}\\ & =\sqrt{2{\sqrt{5}}^2}\\ & =\boxed{\sqrt{10}}.\end{array}$$