Why is dark energy proof for an open universe? Dark energy appears to be increasing at the rate which the universe expands. Why does this show the universe is open?
 A: In this context the term closed means that the universe will recollapse back to a Big Crunch. So if we graph the scale factor $a(t)$ as a function of time we'll get something like:

Don't take the shape of the curve too literally - I've just drawn some random squiggly line. The only thing we're sure of is that it must start at $a=0$ and end at $a=0$ to be a closed universe.
I suspect the question expects you to take dark energy as a cosmological constant. This is widely believed to be the case, but in principle the dark energy density need not be constant but might have some unknown dependence on time. If so then all bets are off as it's impossible to predict how the universe will evolve in the future.
Anyhow, if dark energy is a cosmological constant, $\Lambda$, and if the matter behaves like pressureless dust with some density, $\rho$, (both of which seem reasonable assumptions) then the acceleration of the scale factor is given by the Friedmann equation:
$$ \frac{\ddot{a}}{a} = - \frac{4\pi g \rho}{3} + \frac{\Lambda c^2}{3} $$
The density $\rho \propto a^{-3}$ so for times near the Big Bang $\rho$ can be arbitrarily large and therefore:
$$ \frac{4\pi g \rho}{3} \gt \frac{\Lambda c^2}{3} $$
and therefore $\ddot{a} \lt 0$ i.e. the expansion is decelerating.
But as the universe expands, and $a$ increases, the density decreases so the rate of deceleration decreases. If the increase in $a$ and decrease in $\rho$ ever reaches the point where:
$$ \frac{4\pi g \rho}{3} \lt \frac{\Lambda c^2}{3} $$
Then the acceleration changes sign so we now have $\ddot{a} \gt 0$. Once this happens any further expansion just makes $\ddot{a}$ even more positive and the acceleration increases further. This reduces the density even further and we get a runaway expansion ending eventually with a de Sitter universe undergoing exponential expansion.
So if this is a good description of our universe, and if we observe that $\ddot{a} \gt 0$, it means we are doomed to end up with an open universe because by this point it is impossible for our universe to recollapse again.
A: It is the other way around; from observational data (supernova and CMB), we know the Universe is flat (open but not negatively curved). To satisfy this, we introduce the concept of dark energy. And it doesn't grow at the same rate as the expansion of the Universe as the result of a coincidence. Hypothesizing that vacuum fluctuations give rise to dark energy, one can assume it is a constant of nature. That is, independent of the scale factor of the Universe (or time), a $\text{cm}^3$ of space will always have the same amount of dark energy. Assuming dark energy is the dominant component of the Universe (not actually the case), it will result in an exponential expansion. 
