the professor stated that, even if the input will be the same as the output, one should normally use a voltage regulator regardless.
Nope. A linear voltage regulator can't produce an output voltage higher than the input. In fact, output will always be lower, and the minimum input-to-output voltage where the regulator still regulates is called "dropout voltage". Here's a simplified schematic.
A low dropout regulator is basically a PMOS or PNP transistor acting as a controlled current source. The error amplifier adjusts current to keep the output voltage stable. Since the PMOS has non zero resistance when fully on, there is always a dropout voltage.
If input voltage is too low, for example 5V input on a 5V regulator, then the error amp will turn on the transistor fully in an attempt to bring the output voltage up. Since that's not possible in this case, it just sits there, being useless and wasting a bit of voltage.
There is a type of switching regulator that can produce an output voltage equal, lower or higher than the input: the buck-boost converter. That's what you'd use if you had a battery voltage going from 4.2V to 3V as it discharges, and you absolutely needed 3V3, for example. But in this case, it's not necessary, because the datasheet of your chip says:
This means with a normal USB port delivering close to 5V, you got quite a bit of margin to accomodate voltage drop in the cable, ferrite bead, etc.
What you do need is a decoupling cap on your board, to provide a low impedance power supply at high frequency, but your shematic already has those.
Note that your ferrite bead can ring with the low ESR ceramic capacitors and create a voltage spike when you will plug the USB cable. That can fry your chip. One solution is to add a low cost general purpose aluminium electrolytic cap on the right side of the ferrite bead, 100µF or so, in parallel with the ceramic decoupling caps.
Also, if you need a regulator for something else, AMS1117 is not the best choice. It has a very slow transient response, so if you use it with a load that draws chopped current, like a micro that goes to sleep, output voltage will not recover quick enough when the micro wake up.