Sub-sector covers for disks and balls Define an open $k$-sector of a disk as the portion between two radii separated
by an angle of $2\pi/k$, but open along the two radii (and closed along the
circle boundary).
Call a set a sub-$k$-sector if it is a closed set and a subset of an open $k$-sector.
So an open $2$-sector is a half-disk minus the diameter,
and a sub-$2$-sector is a closed set that fits inside this set.

 


          

Left: Coverage by three sub-$2$-sectors.
Right: Four sub-$4$-sectors. Diameters are dotted lines.


It is clear that one can position three sub-$2$-sectors so that they cover a disk (above, left).


Q1. Is it correct that three sub-semiballs suffice to cover
  a ball in $\mathbb{R}^3$? What is the generalization to $\mathbb{R}^d$?

By a sub-semiball I mean the natural analog: a closed subset of a half-ball
minus its bounding plane.

Q2. Does it require six sub-$4$-sectors to cover a disk?

I would like to believe that $k{+}1$ sub-$k$-sectors 
suffice—-so little of the quadrant is missing!—but
the drawing above suggests otherwise.

Update. Here is my attempt to illustrate Douglas Zare's answer to Q2.

          


 A: I'll address question $2$: Since the distance between $(1,0)$ and $(0,1)$ is $\sqrt{2} \gt 1$, you can cover a neighborhood of a radius with a sub-$4$-sector. The remainder of the disk can be broken into four sub-$4$-sectors by dividing it radially. The same idea lets you cover a disk with four sub-$3$-sectors or six sub-$5$-sectors.
When $k \ge 6$, a sub-$k$-sector has diameter less than the radius of the circle. Any sub-$6$-sector containing the center can't contain a point on the boundary. Any sub-$6$-sector contains less than $2\pi/6$ radians of the boundary, so it takes at least seven to cover the boundary. So, at least eight sub-$6$-sectors are needed to cover the disk. 
This does not prove that it requires $k+2$ sub-$k$-sectors to cover a disk for $k \gt 6$, but that many suffices by covering a neighborhood of a radius using two sub-$k$-sectors, then dividing the remainder radially.
A: On Q1. The intersection of sub-semiball with the boundary sphere of the ball cannot contain two opposite points. Thus, bu the Borsuk--Ulam(--Lusternik--Shnirelman) theorem it is impossible to cover this sphere by three such closed sets.
Let me now finish the consideration of Q2 started by Douglas. For $k\geq 7$, again $k+1$ sub-$k$-sectors suffice!
Take $k-1$ (not sub!) $k$-sectors centered at the center of the disk with very small overlaps. Taking their appropriate sub-$k$-sectors we cover everything except a very small "star" around the center and, roughly speaking, a $(k-\varepsilon)$-sector of a bit larger radius. Now put the $k$th sub-$k$-sector so as to cover the star and most of the $(k-\varepsilon)$-sector, without some part around its boundary arc. This arc, together with its boundary, can now be covered by the last sub-$k$-sector.
You may see an example for $k=7$ below. I tried to do my best, but it seems that one needs to be very precise in order to cover the central star; still, it's obviously possible.

