Using a envelope detector to demodulate low voltage signal

Go with @Wouter van Ooijen's suggestion:

^{simulate this circuit – Schematic created using CircuitLab}

Cut the wiring / trace at either end of the relay coil on the grill. Bridge them with a normally closed reed switch. By experiment find out how many turns of expresso wire you need to get the reed switch to change over and add, say, another 25% to give reliable switching.

No electronics!

The frequency ratio between your carrier and your modulation is only 50:1, which is less than 6 octaves (less than 2 decades). You'll need a much better filter than a single-pole RC network to effectively separate the two.

The output of the diode alone is a half-wave rectified version of the original signal. While the average value of the original signal was 0, the average value of the rectified signal is proportional to the envelope of the signal. It is the job of the output filter to separate this average value from all of the other higher-frequency components.

To eliminate obvious distortion on the oscilloscope, you'll need to attenuate the carrier voltage by a factor of 100 or more relative to the baseband signal, which is a power ratio of 40 dB. To accomplish this within a 50:1 frequency ratio, you'd need a filter with a minimum slope of

$$\frac{40 dB}{\log_{10}(50)} = 24 dB/decade$$

This assumes that the cutoff frequency of the filter is exactly equal to the highest frequency of the baseband signal, which means that that frequency will also be attenuated by 3 dB. The point is, you'll need at least a two-pole filter, and you'll need something more sophisticated than a resistor and capacitor connected directly to the diode, which creates asymmetries in the time-domain response because of its varying source impedance.