If sweep delays compressibility, why does 747 cruise slower than A380?
From Wikipedia, the cruise speeds of the both airplanes are fairly similar:

A380, nominal wing sweep angle 33.5 degrees, cruise speed Mach 0.85

Boeing 747: nominal wing sweep angle 37.5 degrees, cruise speed Mach 0.84 .. 0.9, depending on the model.


So those numbers would contradict the claim in your question, but align with your expectations.
There are several other things to consider. The nominal cruise speed of these planes is more dictated by efficiency, required power for a given speed, and fuel prices. The 747 is an older design, when fuel was cheaper, the general tendancy is that modern airplanes reduced the cruising speed slightly in favor of higher efficiencies.
considering the sweep, raw numbers:

cos(33.5 deg)= 0.83
cos(37.5 deg)= 0.79

So the difference of approx 4% here seems to reflect the difference in cruising mach numbers: the higher sweep of the Boeing 747 reduces apparent Mach number roughly by the same amount that it is flying faster.
Then, for both airplanes, and certainly for the A380, the wing aspect ratio is quite low. So there's a point to challenge how much the sweep actually angles differ, from an aerodynamic point of view. It may be very well that effective angles depend a lot more on local airfoil shapes, low aspect ratios, and other 3d effects than they depend on the nominal sweep angle.
All that said, the swept wing has more advantages than just reducing the apparent Mach number. There is

enhanced aerodynamic stability around the yaw axis (this may even help in the event of asymmetric thrust due to an engine failure)
enhanced mechanical stability of the wing itself under aerodynamic loads
trim properties in transsonic aero regimes

The final design is then a result of many different considerations.
Eg, IIRC, the wing span of the A380 had a limitation so it would &quot;fit into standard airports&quot;, and there were likely some trade-offs in aerodynamics for that reason.