Can Hubble red shift be interpreted as time dilation? Can we interpret the de Sitter universe as a spherical cosmic horizon null surface of finite radius, centered at Earth, and containing the Hubble volume of space where time is dilated and radial dimensions contract closer to the edge in such a way that objects closer to the edge do not recognize that they are radially contracted?
Everything is attracted to the edge, but the total radius remains more or less constant and emits de Sitter radiation at finite temperature.

 A: What you are describing is the static coordinate system for the de Sitter space, in which the metric could be written as
$$
ds^2 = -\left(1-\frac{r^2}{\alpha^2}\right)dt^2 + \left(1-\frac{r^2}{\alpha^2}\right)^{-1}dr^2 + r^2 d\Omega_{2}^2.
$$
This is a static universe (not just 'more or less').
We see that at $r=\alpha$ the metric has a cosmological horizon. The region $r<\alpha$ contains the operationally meaningful portion of the de Sitter space, which can be probed by a single observer located at origin. 
The factor at $dt^2$ could indeed be interpreted as defining position-dependent time dilation. An object held at a fixed distance from the observer appears redshifted. If released, such object will be accelerating toward the horizon, so your description is accurate in this respect.
One point I object to in your description is the 'contraction' of radial dimension. The $dr^2$ part of the metric tells us how we actually measure the radial distance (from  origin):
$$ R = \int\limits _0^r\frac{dr}{\sqrt{1-\frac{r^2}{\alpha^2}}}.$$
At the same time your illustration is qualitatively correct in displaying objects in the $(r,theta)$ coordinates. 
A: Events are always undergoing acceleration as they evolve forward in the time dilated continuum. Therefore, when we look out into space beyond the solar system, and back in time, we are also looking down a time dilation gradient into slower time. The observer’s invariant relative rate of time is always faster than that in frames in the perceived past, and we find that as D → ~13.9 Gly, difference in the rate of time, denoted here as "dRt", → 1 s/s, recessional V → c, and lateral V → 0, just as it does near the event horizon of a black hole. Slower time results in lower frequency and the Hubble shift.
Assuming a Hubble constant of 70 $\frac{\frac{km}{s}}{Mpc}$, we find the apparent recessional velocity reaches c at 4282.7494 Mpc = 13.968062372 Gly.
For a 1s/s dRt at this distance the rate of change is:
1/13968062372 = 7.1592*10^-11 s/s/ly = 2.3349516024*10^-4 s/s/Mpc.
So for each Mpc the $dRt = 2.3349516024 \cdot 10^{-4}$ s/s and:
$c \cdot (1 + dRt) = (299792.458) m/s \cdot ((1+(2.3349516024 \cdot 10^{-4})) s = 299862.458 m$ and:
299862.458 - 299792.458 = 70 km/s/Mpc = the Hubble constant
This indicates that the forward evolution of time includes a universal constant of acceleration.
Because we are always being accelerated forward in the rate of time, and therefore apparently space, events in the past must appear to accelerate away from us in the opposite direction.
This also creates the impression we are at the center of the universe and leading it in its evolution. 
Please also note that the solution works for a difference in the rates of time of exactly 1 s/s. Does any other theory you know of account for a 1 s/s difference in the rates of time between us and 13.9 Gly?
