What the image represents is the energy loss/gain of a whole bunch of electrons that interact one at a time with a light wave traveling along a tiny metal wire (we call this kind of captive light wave a "phonon").
For each electron fired at the wire, it can either pass through unscathed, or it can interact with an electron and scatter off with a different energy, like two billiard balls hitting each other. The collision is fundamentally a collision between two particles. However, the fact that the properties of the collisions vary sinusoidally along the length of the wire imply that the photon is also acting like a wave oscillating up and down the wire.
I disagree that this is the "first time" that we've observed simultaneous wave-particle behavior, you can see the same thing by firing photons at two narrowly-placed slits. This results in a diffraction pattern as if the photons were interfering like waves, but we can confirm that we're detecting single photons at the detector, and you can even confirm that each photon travels through one slit or the other ... See https://en.wikipedia.org/wiki/Double-slit_experiment
Yep, that should have said "plasmon". I messed up all of my comments on this thread. A plasmon isn't technically light either, it's an oscillation of the electric field in a conductor -- but it has wave/particle duality like a photon.
Your eyes don't have quantum energy packet receptors :-) A normal photograph is made by bouncing light off something and detecting it. If you are using a CCD, you are still using false colour. It just happens to correspond to the true colour.
In this experiment, the electrons interact with the photons to produce quantum energy packets, which the microscope detects. What is the colour of the quantum energy packets? The question has no meaning, because they aren't light. Since the colour is arbitrary, they can choose to colour it in any way they find useful.
Having said that, I also don't understand the "photograph".
