This is a numerical calculation of an improved covering like the one proposed by Gerhard Paseman. It
gives the following list of 22 centers for 1/2-balls that cover the unit ball.
All, besides the central one, are on the sphere with radius Sqrt[3]/2.

2 are on the poles

6 on the upper hemisphere at latitude
t = 0.20483559485813116` Pi

6 on the lower hemisphere at latitude
t = 0.20483559485813116` Pi

7 lie distributed over the equator with angular distance
2/7.15 Pi, and phase shift 0.86 wrt the six upper and lower.

I didn't yet calculate all intersections of neighboring balls explicitely.

Centers:
{{0, 0, 0}, {0.570962, 0.651155, 0.}, {-0.137028, 0.855116,
0.}, {-0.745836, 0.440145, 0.}, {-0.81481, -0.293402,
0.}, {-0.294026, -0.814585, 0.}, {0.439574, -0.746173,
0.}, {0.855011, -0.137682, 0.}, {0.599996, 0.346408, 0.519621}, {0.,
0.692816, 0.519621}, {-0.599996, 0.346408,
0.519621}, {-0.599996, -0.346408, 0.519621}, {0., -0.692816,
0.519621}, {0.599996, -0.346408, 0.519621}, {0.599996,
0.346408, -0.519621}, {0., 0.692816, -0.519621}, {-0.599996,
0.346408, -0.519621}, {-0.599996, -0.346408, -0.519621}, {0.,
-0.692816, -0.519621}, {0.599996, -0.346408, -0.519621}, {0., 0.,
0.866025}, {0., 0., -0.866025}}

Graphics:

ddp = 0.86; equator =
Take[Table[{Cos[p], Sin[p], 0}, {p, ddp, 2 Pi, 2/7.15 Pi}], 7];
t = 0.20483559485813116 Pi; dp = 0; up =
Table[{Cos[p] Cos[t], Sin[p] Cos[t], Sin[t]}, {p, Pi/6 + dp,
11/6 Pi + dp, Pi/3}];
dn = Table[{Cos[p] Cos[t], Sin[p] Cos[t], -Sin[t]}, {p, Pi/6 + dp,
11/6 Pi + dp, Pi/3}];
poles = {{0, 0, 1}, {0, 0, -1}};
out = Sqrt[3]/2 Join[equator, up, dn, poles];
Graphics3D[{{Opacity`[1]`

, Red, Sphere[{0, 0, 0}, 1/2]}, {Opacity[0.4],
Red, Sphere[{0, 0, 0}, 1]}, {Opacity[0.2],
Sphere[ #, 1/2] & /@ out}}, Boxed -> True]

[*Graphic from the above code added by J.O'Rourke*:]

KF-PS: I changed the phase shift from 0.85... to ddp = 0.86. In this case numerical calculation shows that in fact the minimum of the maximum intersection points of three neighboring spheres is >1, which implies that the unit ball is covered.

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