I assume that you are OK with parts a and b. Now let's develop an intuition about part c. In the first instance, let's simplify the problem even more. Assume that the tilt is around the horizontal axis. So the normal to the plane makes an angle $\theta$ with the horizontal. The bottom part of the paper (say length $l$) will not change its apparent size when the rectangular piece of paper is tilted. The vertical side of the rectangle, $L$, when tilted an angle $\theta$ will change its apparent size to $L\cos\theta$. The apparent area of the rectangle is now $$dS'=lL\cos\theta=dS\cos\theta$$
Now let's get back to the original statement, by rotating the tilted piece of paper around the axis towards the eye. Notice that the normal to the paper is still at angle $\theta$ with respect to that axis (it moves over the surface of the cone with half angle $\theta$). Also, due to symmetry, the apparent area $dS'$ does not change with this additional rotation. Therefore part c is proved.
In the figure above, initially the piece of paper is vertical (as seen from the side). The length $L$ is along the $OA$ axis, $l'$ is perpendicular to the figure (out of the image). The original normal to the paper points along $B$ (towards the eye). In the first case I've rotated the paper along the axis through $O$, perpendicular to the figure, by an angle $\theta$. That moved $B$ to $B'$ (direction of the normal changed by $\theta$) and $A$ to $A'$ (also by angle $\theta$). To prove that both angles $\angle BOB'$ and $\angle AOA'$ are the same, just use the fact that adding $\angle B'OA$ to both of them yields $90^\circ$. That means that the apparent size of $A'O$ is the vertical projection $A'O\cos\theta$.
For the last part, I took the above figure, and rotated around the $OB$ axis, in such a way that the $OB'$ describes the surface of the cone with vertex at $O$. No matter where on this cone you put $B'$, the angle between $OB'$ and $OB$ is unchanged.