Question: It has been said that our universe is going to eventually become a de Sitter universe. Expansion will accelerate until their relative speed become higher than the speed of light. 

So i want to understand what happens after this point: so from our point of view, we see a progressively shrinking event horizon (each galaxy sees one, arguably each spacetime point sees one). Now, what is the Hawking Radiation expected from this event horizon? It would seems that the event horizon area is shrinking around us, but its actually a black hole turns outwards: the black hole is actually ***outside*** the event horizon, and the visible "well-behaved" spacetime without singularities in the inside of this horizon

In any case, intuitively (i don't have any arguments to support this) i would expect that the Hawking Radiation *inside* the horizon would grew larger as this horizon would shrink, but i would love to hear what is actually expected to happen at this point

Answer: 
We are *already* living in a nearly empty de Sitter space - the cosmological constant already represents 73% of the energy density in the Universe - and the Universe won't experience any qualitative change in the future: the percentage will just approach 100%.

However, once the space may be approximated as an empty de Sitter space, all moments of time are physically equivalent. It's because de Sitter space belongs among the so-called "maximally symmetric spacetimes" - in which each point may be mapped to any other point by a symmetry transformation (isometry).

So nothing will change qualitatively: the radius of the cosmic horizon will converge towards those 100 billion light years or so and never change again; we are not far from that point.

Yes, it is true that de Sitter space is analogous to a black hole except that the interior of the black hole is analogous to the space behind the cosmic horizon - outside the visible Universe. The de Sitter space also emits its thermal radiation, analogous to the Hawking radiation. It's a radiation emitted from the cosmic horizon "inwards". Because the interior of the de Sitter patch is compact, unlike its black hole counterpart (the exterior of the black hole), the radiation is reabsorbed by the cosmic horizon after some time and the de Sitter space no longer loses energy.

While the right theoretical description of the thermal radiation in de Sitter space is a theoretician's puzzle *par excellence* (we are only "pretty sure" about the semiclassical limit, and don't even know whether there exists any description that is more accurate than that), it has absolutely no impact on observable physics because the typical wavelength of the de Sitter thermal radiation is comparable to the radius of the Universe. (Note that it's true for black holes, too: the wavelength of the Hawking radiation is mostly comparable to the black hole radius.)

Such low-energy quanta are obviously unobservable in practice - and in some sense, they're probably unobservable even in theory. You should imagine that there are just $O(1)$ thermal photons emitted by the cosmic horizons inside the visible cosmos whose energy is $10^{-60}$ Planck energies per photon. From an empirical viewpoint, it's ludicrous.
The Hawking radiation of a deSitter space is given, as always, from the near-horizon metric, consistently redshifted to fill all space. Normalizing the cosmological constant appropriately:

$$ ds^2 = - (1- {\Lambda\over 3} r^2) dt^2 + {dr^2 \over (1 - {\Lambda\over 3} r^2)} + r^2 d\Omega^2 $$

If you flip the sign of $dt$, you can identify the metric of a 4-sphere after appropriate coordinate transformations, and from this read off the periodicity of t, which goes around the sphere. But this is not necessary. The near horizon metric is Rindler (as usual for hot horizons) and writing $r=r\_0 - {6\over\Lambda} u^2 $ where $r\_0$ is the deSitter radius, you find:

$$ ds^2 = - ( {\Lambda^2 u^2\over 9}) dt^2 + du^2 $$

Which gives the imaginary time period is $\Lambda u\over 3 $, so that the near horizon temperature is $3\over\Lambda u$. Extending this using the redshift factor, the temperature at the center is

$$ 1\over {2\pi r\_0}$$

Where

$$ r\_0 = \sqrt{3\over\Lambda}$$

Which is the usual deSitter temperature. This temperature is locally the same everwhere, because the space is isotropic. The horizon is static, and stable in equilibrium with this thermal bath, it doesn't grow and it doesn't shrink.