HPF purpose on motor circuit
The graphite rheostat may get a little "noisy" with age due to poor contact at worn positions. Any making and breaking of the contact would have two effects:  


Arcing with further burning of the affected zone. This would be exacerbated by the inductive characteristics of the motor.
High-frequency emissions from the wiring. I would guess that this might be less of a concern but perhaps it dates from the days of medium and long wave radios which would have been more susceptible to low-frequency EMI (as they operate in the high kilo-hertz region).


As you state, the capacitor would act as a high-pass filter (HPF) shorting out the high-frequency components of the noise. The resistor is probably there to discharge the capacitor so that anyone unplugging the device from the mains wouldn't get a tingle from the plug pins if touched. (I had a computer that must have had a small capacitor across the 230 V input. It's quite a surprise to get a jolt from the mains plug pins.)


  I though that HPF's were supposed to block current with low frequencies ...


Correct, and the opposite view that HPFs pass high frequencies.


  ... based on a calculator it stands at 1Hz for the values given. 








simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. (a) A high-pass filter. (b) A different high-pass filter.

You may be confusing the standard high-pass RC filter with the purpose of this circuit. The resistor you should be considering is that of the motor which will be, perhaps, in the order of 100 Ω or so. The corner frequency would then be \$ \frac {1}{2 \pi RC} = \frac {1}{2 \pi 100 \cdot 100n} = 16~kHz \$.


  But if it lets high frequency current pass wouldn't it keep the circuit continuously closed even working?


You don't need to worry about that. The impedance of a 100 nF capacitor is given by \$ Z = \frac {1}{2 \pi fC} \$. At 50 Hz this works out to be \$ \frac {1}{2 \pi 50 \cdot 100n} \approx 32~k \Omega \$ so a tiny current will pass through it.