In your hypothetical example of trying to detect a 3VDC signal against a background of 250kVDC noise (say between the phase 1 and phase 2 power lines), imagine the signal is being measured by a well-insulated lineman who can put his DMM's positive lead on the "phase 1" 250.003kVDC wire, and the negative lead on the "phase 2" 250.000kVDC wire. Now this floating DMM can measure the 3VDC difference between the two 250kV wires. Even though the lineman and his DMM are now at 250kV potential w/respect to ground, his DMM is measuring only the difference between its positive and negative leads. The DMM doesn't blow up, because the entire thing is within an acceptable "common mode" range due to voltage isolation.
(I don't want to get sidetracked on the actual details of AC vs DC HV power distribution systems, just giving an idea how differential measurement works.)
ECG/EKG (Electrocardiogram) has a similar problem to EEG (Electroencephalography): common-mode noise exceeds the differential signal by many orders of magnitude. And because the human body is a high-impedance signal source, incidental power-line noise is a big problem. The signal of interest is a very small voltage (mV), riding on top of a much larger noise voltage (several Volts).
Differential amplifiers are the key to extracting the signal from the noise. Multiple electrodes are used, and the signal is measured in the voltage difference between electrodes. As long as the common-mode input range provides enough dynamic range to accommodate the power-line noise and unwanted EMG (muscle) artifacts (which must be presented equally at both the positive and negative inputs), the small differential signal is amplified to a useful level without saturation.
I don't know about EEG (Electroencephalography) specifically, but ECG (Electrocardiogram) defines its bipolar limb leads (lead I, lead II, lead III) in terms of voltage differences between the various electrodes (LL left arm, RA right arm, LL left leg). Even the so-called unipolar leads are actually differential, as their negative reference (Wilson's Central Terminal) is a virtual ground formed from the average of LL, RA, and LL.
For this differential measurement to be possible, there has to be balance. It's critical that both the positive and negative leads have to pick up exactly the same common-mode noise. Any mismatch between the input leads, will cause mismatched common-mode noise, which is indistinguishable from the differential signal. So physical symmetry is important. And both the positive and negative signals have to be within an acceptable common-mode input range (determined by the op-amp's input common-mode range and the input network impedance).
Although your example of a 250kV power line is hypothetical, in a real ECG system they do have to worry about not only measuring differential signals in the mV range, but they also have to be able to withstand hundreds of volts and many joules from an externally applied defibrillator. If the patient has a heart attack, the doctor won't bother to disconnect the poor ECG machine before applying the defibrillator shock to revive the patient. Anyone and anything connected to the patient, will get shocked by the defibrillator. So at least in the medical electronics world, there does have to be some circuitry to protect against high voltage, yet not mess up the measurement of high-impedance, low-voltage signals of interest. Using symmetrical defibrillator protection circuitry on each electrode input, makes the protection circuit's error voltage contribute to the common-mode noise, which is rejected by the differential amplifier.
Take a peek at figs 7 and 8 in the data sheet. for +/- 5V supply, the common mode input voltage cannot be less than about 4 volts. I suspect that with your 4V supply, and ground as the negative supply, you can't get within 750 mV of ground as a common mode input. While it may operate single supply, it is not rail to rail on the input.
The fact that there is no equivalent to figs 7 and 8 for single supply operation suggests that this is not a very good choice of an instrumentation amp for a single supply emg. I don't know any rail to rail dual package, but suggest you peek at the AD623, which actually works for a common mode voltage a bit BELOW the negative rail.
If your output is stuck at Vref, that would suggest that if it were grounded you'd be stuck at the negative rail. You don't say the size of the sine wave you're using as input, but I suspect it's less than a volt, and below the valid range for common mode input given single supply use. Try putting in a DC 3 volts and see what happens.
Also, that's not how you use a reference terminal on an instrumentation amp. You offset the output stage by providing a LOW IMPEDANCE reference voltage, usually the output of an op amp.
Best Answer
The reference pin has a 20k ohm input impedance and your potential divider may be formed from large resistors i.e. values in excess of 10k - this will cause a problem. Try driving the ref pin with a voltage source. Also note that there is a leakage current from this pin of about 50uA. This will produce a 0.5 volt error when fed from a voltage source with a 10k impedance.