Spacetime can dynamically evolve in a way which apparently violates special relativity. A good example is how galaxies move out with a velocity v = Hd, the Hubble rule, where v = c = Hr_h at the de Sitter horizon (approximately) and the red shift is z = 1. For z > 1 galaxies are frame dragged outwards at a speed greater than light. Similarly an observer entering a black hole passes through the horizon and proceeds inwards at v > c by the frame dragging by radial Killing vectors.
The Alcubierre warp drive is a little spacetime gadget which compresses distances between points of space in a region ahead of the direction of motion and correspondingly expands the distance between points in a leeward region. If distances between points in a forwards region are compressed by a factor of 10 this serves as a “warp factor” which as I remember is $w~=~1~+~ln(c)$, so a compression of 10 is a warp factor 3.3. The effect of this compression is to reduce the effective distance traveled on a frame which is commoved with the so called warp bubble. This compression of space is given by $g_{tt}$ $=~1~-~vf(r)$.
Of course as it turns out this requires exotic matter with $T^{00}~<~0$, which makes it problematic. Universe is also a sort of warp drive, but this is not due to a violation of the weak energy condition $T^{00}~\ge~0$. Inflationary pressure is due to positive energy. The gravity field is due to the quantum vacuum, and this defines an effective stress-energy tensor $T^{ab}$ with components $T^{00}~=~const*\rho$, for $\rho$ energy density, and $T^{ij}~=~const*pu^iu^j$, for $i$ and $j$ running over spatial coordinates $u^i$ velocity and $p$ pressure density. For the de Sitter spacetime the energy density and pressure satisfies a state $p~=~w*\rho$ where $w~=~-1$. So the pressure in effect is what is stretching out space and frame dragging galaxies with it. There is no need for a negative energy density or exotic matter.
Negative energy density or negative mass fields have serious pathologies. Principally since they are due to quantum mechanics the negative eigen-energy states have no lower bound. This then means the vacuum for these fields is unstable and would descend to ever lower energy levels and produce a vast amount of quanta or radiation. I don’t believe this happens. The Alcubierre warp drive then has a serious departure between local laws of physics and global ones, which is not apparent in the universe or de Sitter spacetime. The Alcubierre warp drive is then important as a gadget, along with wormholes as related things, to understand how nature prevents closed timelike curves and related processes.
Addendum:
The question was asked about the redshift factor and the cosmological horizon. This requires a bit more than a comment post. On a stationary coordinate region of the de Sitter spacetime $g_{tt}~=~1~-~\Lambda r^2/3$. This metric term is zero for $r~=~\sqrt{3/\Lambda}$, which is the distance to the cosmological horizon.
The red shift factor can be considered as the expansion of a local volume of space, where photons that enter and leave this “box” can be thought of as a standing wave of photons. The expansion factor is then given by the scale factor for the expansion of the box
$$
z~=~\frac{a(t_0)}{a(t)}~-~1
$$
The dynamics for the scale factor is given by the FLRW metric
$$
\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\rho}{3}
$$
for $k~=~0$. The left hand side is the Hubble factor, which is constant in space but not time. Writing the $\Lambda g_{ab}~=~8\pi GT_{ab}$ as a vacuum energy and $\rho~=~T_{00}$ we get
$$
\Big(\frac{\dot a}{a}\Big)^2~=~H^2~=~\frac{\Lambda}{3}
$$
the evolution of the scale factor with time is then
$$
a(t)~=~\sqrt{3/\Lambda}e^{\sqrt{\Lambda /3}t}.
$$
Hence the ratio is $a(t)/a(t_0)~=~ e^{\sqrt{\Lambda /3}(t-t_0)}$.The expansion is this exponential function, which is Taylor expanded to give to first order the ratio above
$$
a(t)/a(t_0)~\simeq~1~+~H(t_0)(t_0-t)~=~1~+~H(t_0)(d-d_0)/c
$$
which gives the Hubble rule. $z~=~a(t)/a(t_0)~-~1$. It is clear that from the general expression that $a(t)$ can grow to an arbitrarily large value, and so can $z$. On the cosmological horizon for $d~-~d_0~=~r_h~=~\sqrt{3/\Lambda}$ we have $z~=~1$.
Looking beyond the cosmological horizon $r_h~\simeq~10^{10}$ly is similar to an observer in a black hole looking outside to the exterior world outside the black hole horizon. People get confused into thinking the cosmological horizon is a black membrane similar to that on a black hole. Anything which we do observe beyond the horizon we can never send a signal to, just as a person in a black hole can see the exterior world and can never send a message out.
Best Answer
Currently, there is no known mechanism to getting the warp bubble moving at all, let alone to superluminal velocities. In Alcubierre's original metric, he assumed that the warp bubble would already be moving (see original paper), and didn't consider how that warp bubble would start moving in the first place.
And firing conventional rocket engines or any other propulsion system in a warp bubble doesn't do anything to this effect either. The "inside" of a warp bubble is flat spacetime, and so firing rockets inside of it would be essentially like firing rockets in interstellar space; it doesn't affect the warp bubble itself.
In fact, this is probably the big issue affecting all warp drives that still remains unresolved. Here is another physics SE post that elaborates more on this issue.