Way to proof that you don’t see the cut-off frequency in the plot

The -3dB point is your cutoff frequency. It's just standard practice to define it that way. In order to find what your values should be, I'd go with equal element implementation (it's simpler, and you can correct for gain with a simple gain stage later if you need to). Choose R1=R2, C1=C2, and pick a value for either R or C. This yields the following formula for the cutoff frequency: $$f_0=\frac{1}{2\pi RC}$$ I generally choose a value of C initially, as it's easier to find or make a resistor with a strange value, whereas it's more difficult with capacitors. So, set your cutoff frequency equal to f0, and solve for R.

Here's an example: let's say I want a LPF with f0=250 Hz. I'll choose C to be 0.1 micro and solve for R. $$250=\frac{1}{2\pi RC} \rightarrow 250=\frac{1}{2\pi R(0.1x10^{-6})}\rightarrow R\approx6400\Omega.$$

From there, all you need to do is implement your circuit. Once you know what your value for R is supposed to be, you can use a dual-channel potentiometer that has the correct resistance within it's range in place of the two resistors (for the above example, something like a 10k ohm potentiometer would do the trick). This will allow you to change your cutoff frequency, since it's based upon both R and C.

Edit: As Matt Young suggested in the comments, adding a resistor in series with the potentiometer will set the maximum cutoff, and prevent shorts. It's an excellent addition to the circuit, and will keep some sanity when adding the potentiometers.

First, calling it the "cutoff frequency" leads to misconceptions. "Rolloff frequency" is a better name that gives you a more accurate mental picture of what is really happening.

Using the -3 dB point is not arbitrary. It falls out from the math naturally. For a R-C filter:

  ω = 1 / RC

where ω is the frequency in radians/second, R is in Ohms, and C in Farads. For the frequency in Hz, use:

  f = 1 / 2\$\pi\$RC

If you plot the Log(amplitude) as a function of Log(frequency), such as in a Bode plot, then the -3 dB frequency is where the asymptotes for the pass band and stop band meet. Put another way, at frequencies well into the pass band, the filter looks like a horizontal line. At frequencies well into the stop band, the filter is a line with a slope of 20 dB per decade (+ or - depending on high or low pass). If you draw those two lines and extend them to where they meet, it will be at the -3 dB rolloff frequency.