If, in local system, an accelerated system is completely physically equivalent to a system inside a gravitational field.
Acceleration results in an increase in speed. Right?
What happens when the accelerated "elevator" accelerates to the point where its speed hits (or exceeds?) the speed of light?
Does it stop at the speed of light?
Is that a possible way to distinguish between the two different forces? Or does something else happen?
Accelerating a body beyond $c$ with a force
To accelerate the body you can use a force $\vec{F}$ and the Newton's equation in your lab frame (for experts: the very non-covariant form with relativistic mass $m$): $$\vec{F} = \frac{dp}{dt} = m \vec{a} + \frac{dm}{dt} \vec{v}$$ But for a body with rest mass $m_0$ (i.e. the one measured at rest) and velocity $v$ we have $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ So when the speed approaches $c$, $1-v^2/c^2$ gets very small and $m$ very big. As we get closer and closer to $c$, the mass $m$ grows into infinity as well as the force $\vec{F}$ needed for further acceleration. I.e. you would need an infinite force (and infinite energy) to accelerate a body even to $c$, so breaching the limit by a usual force is impossible.
The more energy you put into a body, the more massive and thus more repulsive to acceleration it will be and you can asymptotically approach $c$ but will never do so.
Gravitational free fall "beyond $c$"
In general relativity you should ask: accelerates beyond the speed of light with respect to who? If the answer is an observer far away from the gravitating body, the moment corresponding to beyond-speed-of-light transition corresponds to the transition beyond the event horizon of a black hole.
There is a simple argument to get an approximate intuition for this. Consider a point far away ("at infinity") from the gravitating body from which the massive particle (or "elevator") accelerates just beyond the speed of light at a height $h$ above the body. Stopping the particle and reversing the process to send this particle back to infinity would mean we have to accelerate it to beyond speed of light. Which is impossible, so no physics leave from $h$ or lower. The observer far away would just see a black void from $h$ and lower. This is an approximate notion of a black hole.
But the whole argument is very inexact. It is more accurate to talk about general relativity, "space-like" and "time-like" directions etc. And surprisingly enough, it turns out that such a "censorship" $h$ is universal for all particles shot at any speed from any observer outside the black hole. But it is the same $h$ only in the sense that not even a particle of speed $c$ can escape towards them.
So a usual force cannot reach this situation and gravity makes sure it gets censored. In a sense you could say the particle beyond the horizon is moving beyond the speed of light with respect to an observer outside, but there is no observational sense to this. Most importantly, special relativity is never locally violated. Only when formally comparing far away causally disconnected events we can seem to violate it (but it is not true).
Your accelerated elevator will never reach the speed of light, but it can always accelerate more, gaining a finite amount of energy but only infinitesimally more speed as measured in the original rest frame.
Here's a thought experiment I find useful. Imagine you are standing at rest, and see elevators A, B, and C pass by going 0.99c, 0.999c, and 0.9999c, respectively. You strap on your even-more-awesome-than-elevators jetpack and accelerate until you catch up with elevator A. You now find, in your new frame of reference, that elevator B is still going 0.82c! Elevator C is still traveling around 0.98c! If you caught up with elevator B, elevator C would still be going about 0.82c.
There's always more room at the top.
The forces are not totally equivalent. They are only equivalent from the perspective of a point particle. Standing in an accelerating elevator in empty space, every part of your body feels the same force. Standing in the gravitational field of a planet, a distance $d$ from the center, your head and your feet feel forces which differ by $$ \delta F = G M m\left(\frac{1}{(d+h)^2} - \frac{1}{d^2} \right) = \frac{G M m}{d^2}\left( 1 - 2\frac{h}{d} + O(2) -1 \right) \approx \frac{-2 G M m h}{d^3}. $$
If, in local system, an accelerated system is completely physically equivalent to a system inside a gravitational field.
Yes, but local means in the vicinity of a spacetime point. If you travel too far from your spacetime point you'll find that the curvature is measurable, but this applies to travelling in time as well as travelling in space.
Your question asks what happens as the velocity increases over time, and the answer is that when you move more than an infinitesimal distance in time away from your starting point you are no longer making a local measurement. So you shouldn't be surprised to find that your uniformly accelerated system is no longer equivalent to a gravitational field or vice versa.
