Why does destructive interference not stop a wave? In this picture, the two waves keep moving even after undergoing destructive interference.

 A: The state of the rope is defined not just by the vertical displacement along its length, but also by the vertical velocity.  In this case, the displacement is zero, but the velocity is not.  The parts of the rope are located at their equilibrium positions, but they are still in motion; by Newton's First Law, they will continue to move.  This continued motion regenerates the left- and right-moving wave packets.
For another example, think of a hammer striking a piano string.  The hammer does not initially displace the string, but it sets part of the string in motion.  (This is in contrast to a harpsichord, which excites the strings by displacing/plucking them.)  Although the string starts out, immediately after it has been struck, in its resting position, it has a nonzero perpendicular velocity, which causes it to vibrate.
A: As a complement to the correct answer by Buzz, let me explain why the picture can indeed trick you: this is because it is a collection of snapshots of the physical state of the rope in position space. It completely ignores the other part of the system phase space, the momentum space, which is the part about motions.
When we take a picture (or, as here, when we draw a picture similar in spirit to a photograph) we record all positions taken by the system during the exposure time (the shutter speed). To have a sharp picture, we need the exposure time to be very short, and in the case of a drawing it is actually zero. This amounts to completely killing the dynamics of the system in the record.
Yet, strangely, it seems natural to us that a still picture is a faithfull representation of a system. 
It is interesting to note that an actual photograph taken with a longer exposure time would actually display more of what is happening, in the form of motion blur. In that case, the (c) picture would show that the rope is blurred where the pulses destructively interfere.
A: There are many simulations which show what happens when pulses on a string collide.

with some more interactive than others and the simulation created using GeoGebra is particularly good as so many factors can be controlled.

It also shows the individual pulses during the time they collide.

So what you have here are graphs of displacement of a particle from its equilibrium position $y$ against equilibrium position of the particle $x$ from an origin at a given time $t$.
It is a snapshot of the pulses and these snapshots are called wave profiles.
Indirectly what these graphs show is the potential energy stored in the pulses as the potential energy is related to the displacement of the particles.
What is missing is the other component of energy in the pulses and that is the kinetic energy.
As has been pointed out elsewhere the motion of a pulse can be "seen" by taking a photograph or a still from a video and the blurring is an indication of motion which in this case is a pulse on a slinky.

What you "see" here is the slinky hardly moving at all in the middle of the pulse and at ends of the pulse and the slinky moving with maximum speed at about a quarter of the way from the peak.
A standard way of producing such animations is to use a pulse with an equation of the form $$y_{\rm right}=Ae^{-(kx-\omega t)^2}$$ for a right moving pulse and $$y_{\rm left}=Be^{-(kx+\omega t)^2}$$ for a left moving pulse where $A$ and $B$ are the amplitude of the pulses, $k = \frac{2\pi}{\rm wavelength}$ and $\omega = 2\pi \rm \,frequency$.
To answer the OP what is missing from the GeoGebra simulation are graphs of velocity against position at a given time.
To simplify the Maths I have made all constant $A,B,k,\omega$ equal to one.
A particle at position $x$ has displacement $y_{\rm right}$ (blue graph) at a time $t$ of $$y_{\rm right} = e^{-(x-t)^2}$$ and the particles velocity at a time $t$ and position $x$ (green graph) is $$v_{\rm right} = \left( \dfrac {\partial y_{\rm right}}{\partial t}\right)_x =2(x-t)\,e^{-(x-t)^2}$$.
This is illustrated below for a single pulse travelling to the right at time $t=0$.

Now start with right and left travelling pulses either side of the origin with the velocity of the particles shown in green and red.

and have them collide.

So here the displacement is zero which means that there is no potential energy but the velocity graph (magenta) shows that energy has not been lost but just transferred from potential energy to kinetic energy.

It would be nice if somebody could produce a GeoGebra simulation which also includes the velocity of the particles?
