How can you prove that the angle at which a bullet has to be shot to get the farthest is 45 degrees?

Answer

First, this is only true in vacuum, when there is no air resistance whatsoever.
So, in a vacuum:
We fire a bullet at velocity Vo, angled alpha. The velocity component in the horizontal direction is then Vo cos(alpha), and in the vertical direction is Vo sin(alpha).
Gravity works purely vertically. In the horizontal direction there is no force acting on the ball, and therefore no acceleration. So the speed is constant. So the derivative path in horizontal direction is:

x = Vo cos(alpha) . t

In vertical direction this becomes:

y = Vo sin(alpha) . t – 0.5 gt²

We chose the origin of the axes where the bullet is fired, the x-axis horizontally below the trajectory of the bullet, and the y-axis vertically upwards.

The ball hits the ground when y is zero again, so at time t = 2 Vo sin(alpha) / g

If we fill this in x, we know how far the bullet has hit:

x = Vo² 2 sin(alpha) cos(alpha) / g = Vo² sin(2 alpha) / g

This distance is maximum when the sine is 1 (its maximum value). So 2alfa must be 90° or alpha = 45°