Yes, if a particle would be travelling faster than light, it would always travel faster than light. This is what's called a tachyon, and they have in some sense imaginary mass.
The three regimes, time-like, light-like and space-like (i.e. subluminal, luminal and superluminal space-time distances) are invariant under Lorentz transformation. Therefore anything on a super-luminal 'mass-shell' would always stay there and could not be decelerated to light/ or sub-light speed.
The problem is not that it would violate relativity, but rather causality, since with faster than light information propagation one could 'travel back in time', therefore leading to paradoxes.
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If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $\Delta c/c$ of $3 \times 10^{-4}$.
The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about:
$$ \frac{\Delta c}{c} = 3 \times 10^{-4} \, P $$
In high vacuum labs we can, without too much effort, get to $10^{-10}$ torr and this is around $10^{-13}$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error:
$$ \frac{\Delta c}{c} \approx 3 \times 10^{-17} $$
And this is already smaller than the experimental errors in the measurement.
So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible.
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"Spatially flat" is not the same thing as "flat spacetime". Even if the geometry at any moment of cosmological time $t$ is plain old Euclidean space, the expansion and contraction of this space as time goes on leads to non-zero spacetime curvature.
By way of analogy: the surface of the Earth can be thought of as the union of a bunch of circular lines of latitude, from 90°N to 90°S. Mathematically, a circle has no intrinsic curvature.* Yet the individual circles (with no curvature) form a two-dimensional surface with curvature.
*It has extrinsic curvature when you draw it on a plane, but extrinsic curvature isn't what we're talking about when we talk about spacetime curvature.