This is a problem out of Elementary Differential Geometry by Barrett O'Neill (Chapter 6 Section 3 Number 2).
Let $M$ be a flat surface in $\mathbb{R}^3$ with principal curvatures $k_1$ and $k_2$, where $k_1 = 0$ and $k_2$ is never zero. Want to show that the principal curves of $k_1$ are line segments in $\mathbb{R}^3$.
There is a hint: With $\{E_i\}$ principal and $\alpha''= \nabla_{E_1}E_1$ and use the connection equations.
So far I have concluded that $\alpha'= E_1$, $\alpha ''=E_2$. So, $$\frac{S(\alpha')\cdot \alpha'}{\alpha' \cdot \alpha'}=\frac{\alpha'' \cdot U}{\alpha' \cdot \alpha'}=\frac{E_2 \cdot E_3}{E_1 \cdot E_1}=0.$$ And $S(\alpha') \cdot \alpha' = 0$ implies $\alpha$ is asymptotic. Which means $\alpha$ is not bending away from the tangent plane. Which I suppose I already knew because $k_1$ is $0$. Also I know that the Gaussian curvature is $0$ so my surface is either a plane or trough-shaped. But $k_2$ is non-zero so $M$ cannot be a plane.
I'm not sure how any of this helps me, nor do I know how to use the hint. Also any pointers on why this is intuitively true. I have been looking at examples but cannot figure out why this has to be true.
You're wrong about "trough-shaped." There are cylinders, cones, and so-called tangent developables, all of which are flat, ruled surfaces.
As a hint, I'll suggest that you want to use the Mainardi-Codazzi equations.