Could someone walk me through this proof that the helicoid is the only minimal surface (besides the plane) that is also ruled? 


$\bullet$ How does he get from "the mean curvature is $0$" to "$W$ is covered by two families of asymptotic curves which intersect orthogonally"?
$\bullet$ Why are the rulings asymptotic curves and why does that mean we can choose a point $q \in W$ such that the asymptotic curve, other than the ruling, passing through $q$ has nonzero torsion at $q$?
$\bullet$ How is it that the osculating plane of an asymptotic curve is the tangent plane to the surface and why does that imply that there is a neighborhood $V \subset W$ such that the rulings of $V$ are principal normals to the family of twisted asymptotic curves?
$\bullet$ I know what asymptotic curves are, but what does he mean by twisted asymptotic curves? 
 A: 1) The mean curvature is zero. Therefor the two principal curvatures are one opposit of the other$\pm k$. At a fixed point $P$, the curvature of a normal section in a direction making an angle $\theta$ with one  principal driection is given by (Euler's formula) $k\cos^2\theta-k\sin^2\theta$. From this you easily get two tangent fields, one orthogonal to the other, each describing an asymptotic direction. Integrating these you get the two familes of orthogonal asymptotic curves.
2) the lines of the ruling are asymptotic curves since their curvature, which is a normal sectional curvature, is everywhere zero. Therefore one of this family of curves is indeed a family of lines.
3) If the other family of asymptotic curves was given by curves with torsion everywhere equal to zero they will be plane curves, and thus, since their normal curvature is zero, they will be lines as well. But if this is true everywhere than the surface would be a plane. Thus there is a point in which one of the two asymptotic curve has non zero torsion. to distinguish this curve we will call it a twisted asymptotic curve. 
4) For any curve on a surface the tangent vector is tangent to the surface. If the curve is asymptotic then you can show that its normal vector is tangent to the surface as well. You can in fact show that the normal curvature is proportional to the scalar product between the normal vector to the surface and the second derivative of the curve. Since the osculating plane is parallel to the plane spanned by first and second derivative of the curve and since both its vectors are tangent to the surface then the two planes coincide.
It was written a bit in a hurry so some passages may need to be completed a bit but I think the essentials are here,
