Calibrate without a known good voltage source
If you're serious, somewhere down the line you're going to have to buy - or build - something which is more accurate than the device you're calibrating/testing.

If it were me, I'd build, and I'd use an array of Zeners - LM4040s - rather than a single reference and a voltage divider to get various output voltages from - let's call it - the voltage calibrator.  That way, there's only a single non-precision current limiting resistor required between the raw supply and each Zener's cathode, and the inherent accuracy of each Zener's output voltage will be maintained since it won't be compromised by the tolerance of the voltage divider's resistor chain.

There are 6 individual output voltages available from LM4040s: 2.5V, 3V, 4.096V, 5V, 8.192V, and 10 volts, and they're all wired as shunt regulators, like this,


where the value of Rs is determined by Ohm's law and is:

$$ R_{S} = \frac{V_{DD} - V_{OUT}}{I_Z} $$ 

The LM4040 exhibits a very stable Vout for a very wide range of Iz, so for an LM4040-10 with a Vdd of 12V and an Iz of 1 milliampere, Rs would be:

$$ R_{S} = \frac{12V - 10V}{0.001A} = 2000 \text { ohms}  $$ 

It's important to realize that current taken by the load connected between Vout and ground will cause the current through the Zener to be decreased by that amount, but with high-resistance loads, (voltmeters and such) that change in current through the Zener should have only a miniscule effect on Vout.

Finally, checking DigiKey for prices yields USD 1.78 for a 10 volt +/- 0.1% unit, which is pretty close to dirt cheap, so putting together an array of all six Zeners certainly won't break the bank, especially since the current limiting resistors can all be vanilla 5% carbon films.