How can we predict a Big Crunch when all galaxies are moving further apart? I've read that the further away a Galaxy is from us the faster it moves away. 

By this logic how can scientists predict that there will eventually be a big crunch when every piece of matter is seemingly getting further and further apart?
In a homogeneous and isotropic Universe (even if recent observations challenge this hypothesis), you can derive the Friedmann equations, which describe the evolution of the Hubble constant with time:
$\frac{\dot{a}}{a} = H(t) = \frac{8 \pi G}{3}\rho - \frac{k}{a^2} + \frac{\Lambda}{3}$   (with $c=1$) (Equation $1$)

where $a=a(t)$ is the scale factor, $\dot{a}$ its derivative, $G$ the gravitational constant, $\rho$ the matter density, $\frac{k}{a^2}$ the spatial curvature (a parameter that describes the metric of the Universe), and $\Lambda$ the cosmological constant (an integration constant added by Einstein).
It could be useful to rewrite the equation as:

$H^2 = \frac{8 \pi G}{3}(\rho + \rho_{\Lambda}) - \frac{k}{a^2}$

where $\rho_{\Lambda} = \frac{\Lambda}{8 \pi G}$ is the "density of cosmological constant".

We can also expand the matter density as $\rho = \rho_{matter} + \rho_{radiation}$.

So we have a "total" density $\rho_{tot} = \rho_{matter} + \rho_{radiation} + \rho_{\Lambda}$. The destiny of the Universe depends on this amount.

In case of $\rho_{tot} &gt; \rho_{crit}$, or equivalently a closed Universe ($k=+1$), the equation $(1)$ becomes:

$\dot a^2 = \frac{8 \pi G}{3}\rho a^2 -1$

Which points out that the scale factor must have an upper limit $a_{max}$ ($\dot a^2$ must be positive).
This in turn means that the second derivative $\ddot a$ of the scale factor must be negative, when approaching $a_{max}$, that is the scale factor function inverts its behavior:


Look at here and here if you want to go deeper.

@Bardathehobo This figure shows what I mean when I say that a currently accelerating Universe can still crunch. This is because we are basically ignorant upon the dark energy issue.