How can there be voltage across resistors in series when circuit is open?
Why is it not the case that having one resistor (light bulb in this
case) break does not cause the voltage across all the other
resistors in series to be zero?

It does. There is practically no voltage across the other bulbs. However there is a voltage coming from the mains, which goes through the bulbs and appears at the last one before the disconnection.
The question is, how was he able to detect this voltage without connecting anything to it? The answer is that his detector is able to sense a very small current caused by nearby voltage changes, due to the tiny capacitance between the wire and the detector.
The circuit looks like this:-



simulate this circuit &ndash; Schematic created using CircuitLab
With a mains voltage of 120 V the voltage at the wire is tracing a sine wave going from ~ -170 V to +170 V. This induces a current in nearby metallic objects because the electrons in the wire produce an electric field which repels other electrons, while positive charge attracts them. As the field changes polarity the electrons in the nearby metal are alternately pushed and pulled, creating a tiny current and voltage in time with the AC mains.
As distance increases the electric field and induced current get smaller, but with 170 V at the source a useful voltage still can be detected from a distance of a few cm if the sensor has a very high input impedance. However it will not be able to accurately measure the voltage at the wire, because the loss varies with distance and amount of 'plate' area between the wire and detector, as well as how well the sensor is 'grounded' through the person holding it.