Assumption 1. The voltage will never drop over a long distance (200 feet) because the wire is too large. No matter of low voltage is and how little amps source provide.
No. This is not true. Any wire, on any distance will have a resistance, thus dropping voltage as the effect of this resistance.
This voltage drop can be very important in case of low voltage and high current (lots of amperes) used by the device connected.
Or it can be minimal in the case of very low current used by the connected device or high voltage or both combined.
How much current the supply can provide is not related to the voltage drop. Only the current used by powered device, and the length and the gauge of the cable will define how much voltage you lose.
So this assumption contains two mistakes.
Assumption 2. The voltage will drop more over a long distance (200 feet) on too large wire (AWG 16 vs AWG 6), if the voltage is low (5V) @ 0.5 Amps source providers.
As already mentioned in the comments, AWG 16 is thinner and AWG 6 is thicker, So you have to know what you mean by "large".
In my definition AWG 6 is larger than 16.
So this statement is half correct: An AWG 16 will drop more voltage than AWG 6. But not because it's too large but because it's too thin.
Larger wires present lower voltage drop.
It's also true that 5V, 0.5A over 200 feet is not appropriate. It's better to use a higher voltage, 15V, 24V or 36V, whatever, with a normal wire, and then reconvert the current to 5V with a DC/DC converter. 12V is already not enough.
Now if it's just a 5V signal with minimal current, say, 0.02 mA, then it's OK. But here you already have to know what you are doing.