Electronic – Attempt to use 2n2222 as a switch, but it seems to work below 5v (all the way down to < 1v). Why

Your second circuit is perfect, provided that the pullup on the input is enabled.

When the switch is open, the anode of D1 is at V_DD, and the cathode of D1 is at 13V-V_f. The diode prevents the higher voltage from reaching the input pin and destroying the MCU.

When the switch is closed, the cathode of D1 is at 0V, pulling the anode down to 0.7V and registering a low at the input.

You don't need the resistor since the pullup will not source enough current to damage the diode, but you must have the diode in order to prevent the higher voltage from reaching the MCU.

It is conventional to draw circuits so signals flow from left to right, like so:

^{simulate this circuit – Schematic created using CircuitLab}

(You need a base resistor, R3, to limit base current, and a resistor from base to ground to ensure that the transistor does turn off when the switch is open)

With the switch open, the base will be held at ground by R4, so no current will flow through the transistor, so the collector will rise to a voltage determined by R1, R2, and the 2 volts or so voltage drop in the LED - this will allow current to flow through the LED and light it.

With the switch closed, the base will be pulled up, allowing current to flow through the transistor. If the resistors are selected correctly the transistor will be saturated, pulling the collector down to about 0.2 volts. As the LED requires about 2 volts across it, it will not pass any current.

I really dislike the sentence "Current always takes the path of least resistance", as many beginners seem to read it as "Current takes only the path of least resistance". In face, an electric current takes all possible paths, withthe lower resistance paths passing higher currents than the higher resistance paths.