Special relativity, acceleration and being in two places at once (How to resolve this paradox?): Imagine I am on Earth I have a clock which measures time $t$.  When my Earth clock reads time $t$, then when I look at a moving spacecraft's clock I see the time $t'$.  Let us assume time is measured in years.
Let us suppose the clocks were initially synchronized so that when $t=0$, $t'= 0$.  
Now let us suppose that the speed of the rocket continues to increase in the following way  (Note that I have increased the speed in discrete time steps for convenience, but I'm sure a continuous speed function could be obtained but just mathematically tedious):
At $t = 0$, $v = c\sqrt{3/4}$.  So when $t = 1$, $\delta t'= \frac{\delta t}{\gamma} = 0.5.$

At $t = 1$, $v = c\sqrt{15/16}$.  So when $t = 2$, $\delta t'= \frac{\delta t}{\gamma} = 0.25.$

At $t = 2$, $v = c\sqrt{63/64}$.  So when $t = 2$, $\delta t'= \frac{\delta t}{\gamma} = 0.125.$
and so on...
If $v$ continues to increase over time in this way, then $t'$ will approach 1, but will never reach 1.  It appears from the perspective of the Earthling, that the person in the rocket never ages by over 1 year.  From the person in the rocket however, he will experience time as normal.  
Now let us assume that after the person in the rocket has aged 10 years, he leaves his rocket ship, and travels back to Earth, leaving his rocket to continue accelerating in the same pattern.  Now when the man reaches Earth, will he not see himself in the rocket, where he sees himself not having aged 1 year?
(Ignore optical effects for the purpose of this question.)
 A: Your example can't be usefully described because when you change speed you are changing your inertial frame but you are not specifying how you change frames i.e. what acceleration you use.
Special relativity is perfectly capable of describing accelerated motion. See for example John Baez's article on the relativistic rocket, which gives for constant acceleration:
$$ t_{earth} = \frac{c}{a} sinh\left(\frac{a \space t_{rocket}}{c}\right) $$
where $a$ is the acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket. Since $sinh(x)$ remains finite for all finite $x$, constant acceleration will never give the effect that you want, i.e. for $t_{earth}$ to become infinite at some finite value of $t_{rocket}$. The only way you can achieve this is for the acceleration of the rocket to become infinite, and obviously this isn't physically possible.
However there is an analogy that is close to your question. If you throw someone into a black hole then they will measure a finite time to reach the event horizon while you will never see them reach the horizon i.e. you would have to measure an infinite time before they reach the horizon. The link to your rocket is that the acceleration a shell observer measures for the falling object tends to infinity as the shell observer approaches the event horizon. You can think of this infinite acceleration giving an infinite time dilation just as infinite acceleration would for an accelerating rocket.
But back to the point of your question: although the infalling observer measures a finite time to reach the event horizon, once they've done so an infinite time has passed for the Schwarzschild observers watching them fall. So there is no outside universe to return to. Not that you can return of course, because once you've reached the event horizon all timelike paths lead to the singularity and your inevitable doom.
