why does passive RC filter create THD?
I'll paraphrase what Dave Tweed says. 

If you input a sinewave then there is no misshaping of the signal just an attenuation that is dependent on how high the input frequency is. If you input a square wave, the output will be misshaped because the harmonics (that comprise the square wave) will be attenuated at significantly different levels and you'll likely start to see an output that more resembles a sine wave or triangular wave.

Here's what a 1kHz square wave looks like when it is filtered with the same circuit as the OP has in his question: -



The input is 1Vp-p centred on 3V dc and as you can see the square wave is nearly triangular.