Electronic – Buck converter, duty cycle has inverse relationship with Vout

When Vo is not the correct value (Ierr != 0) the duty cycle has to be adjusted. If Vo approaches Vref the adjustments become smaller until there's the point where Vo = Vref, and no adjustments are necessary anymore. If you would change the duty cycle to 50% at that point, Vo will no longer be equal to Vref.
Besides, if your duty cycle would always be 50% for the correct Vo, then there wouldn't be any regulation, would it? You would just set it to 50, and that would be that.

With the resistor in series with the inductor and your scope across the LED you will get a rising voltage with rising duty cycle then the LED starts conducting at about 2 or 3 volts and you won't see any increase in voltage rise. Try putting the resistor in series with the LED and measuring across R and LED not just across the LED: -

As you can see, to raise the voltage across the LED to about 1.6 volts requires virtually zero current from the MCU pin. To raise it to 2volts requires 20mA and that will be around the limit from your device. Hey, even if it could supply 50mA the LED voltage would rise only to about 2.37 volts. See this if you want more clarification.

Also, make sure your MCU output is capable of handling the kick-back from the inductor without causing problems. Normally a fly-back diode is used or a heavier-duty push-pull stage.

Also, get rid of those meaningless scope images - we know they'll be showing a dc voltage because of the 220uF cap - just tell us what the voltage is by measuring it with a multimeter.