The problem is in a misunderstanding of "simultaneous".
"Simultaneous" refers to two different events that occur at the same time in some particular reference frame, but you're applying it to the same event in two different frames. So it doesn't make sense to say "Pulse has to occur simultaneously for both BoxGuy and PlatGirl." That's a single event - it can't be simultaneous all by itself, even when observed by two different people.
You could, if you want, set the origins of the coordinate systems they are using so PlatGirl and BoxGuy assign the same time coordinate to Pulse. If you do, they will not assign the same time coordinate to Reflect. The time between the events Pulse and Reflect is different in different frames.
Additionally, PlatGirl and BoxGuy will not agree on the length of the boxcar. Your calculation assumes they both measure the length to be $d$, but actually PlatGirl will observe the boxcar to be Lorentz-contracted.
One way to analyze your scenario is to set up coordinate systems $S$ for the boxcar and $S'$ for the platform. We set (x,t) = (0,0) = Pulse in both systems.
In frame $S$ (box), the coordinates are:
Pulse: (0,0)
Reflect: (d,d/c)
Return: (0,2d/c)
In frame $S'$ (platform), the coordinates are:
Pulse: (0,0)
Reflect: $(\sqrt{3}d,\sqrt{3}d/c)$
Return: $(\frac{2\sqrt{3}}{3} d, \frac{4\sqrt{3}}{3} d/c)$
You can verify that in both frames, light moves outward at speed $c$ and returns at speed $-c$
In reply to your edit, yes the durations from Pulse to Reflect and Reflect to Return are the same for BoxGuy and different for PlatGirl. That is just a fact. That's how it is. Notice, though, that the spatial separations are also different. For BoxGuy, these events the same distance apart. For PlatGirl, they are different distances apart. What's the same between frames is the interval $\Delta x^2 - \Delta t^2$.
There have been countless studies on the topic. The easiest for me to understand are the interferometry studies. In these studies, you shine a laser at a semi silvered mirror which splits the beam into two beams at right angles. These each bounce off a mirror and recombine, interfering with each other. You can then move this apparatus in one axis and not the other or rotate it and show that the interference pattern does not change (intuition, such as that gained by thinking about throwing a ball on a train, leads to a different answer).
The most famous of these was the Michaelson-Morley experiment, which showed that the speed of light did not vary with direction or velocity by using the earth's own motion and rotation.
These interference experiments could be though of a a single observer showing that difference in reference frame velocities do not change the velocity of light. If you want it proven with two observers, it may be easier to instead focus on proving that time dilation occurs. That is another expected outcome of relativity. For that, you need to go no further than your local GPS receiver in your phone. Your phone actually has to account for relativistic effects such as time dilation when it is calculating your location. It can be shown that, without that correction, your GPS would be more inaccurate than it is today!
Best Answer
How do you "prove" that 5-3=2? Do the "check your work" operation: final result taken with the reverse operation gets you to the starting point-- 2+3=5.$\checkmark$
The same exercise is done with the Lorentz transformation as a pedagogical tool. If the constancy of the speed of light for all observers leads to the Lorentz transformation, then the Lorentz transformation on a speed of light object should yield a constant speed. And it does. It doesn't prove that the speed of light is constant. It simply shows that the transformation is consistent with the starting axiom.
By the way, "check your work" is an important part of problem solving whether analyzing projectile motion or modeling cosmological expansion: are my solutions consistent with my starting conditions.