I am designing the Vishay VCNL4010 proximity sensor into one of my PCBs. This sensor has an interrupt pin that I don't need to use. Is it OK to just leave this pin floating ("unconnected"), or will this cause problems somehow?
Normally I'd say it won't be a problem: it's an output pin, so it won't be floating as the device itself will drive it.
However, if we look at the datasheet, page 5, first note:
The interrupt pin is an open drain output. The needed pull-up resistor may be connected to the same supply voltage as the application controller and the pull-up resistors at SDA/SCL. Proposed value R2 [a pull-up for the INT pin] should be >1 kΩ , e.g. 10 kΩ to 100 kΩ.
Proposed value for R3 and R4, e.g. 2.2 kΩ to 4.7 kΩ, depend also on the I2C bus speed. For detailed description about set-up and use of the interrupt as well as more application related information see AN: “Designing VCNL3020 into an Application”.
If we look at this application note, Designing VCNL3020 into an Application, page 2:
The SCL and SDA as well as the interrupt lines need pull-up resistors.
I suppose that this is only needed when this line is actually used. However, it's always good to follow the datasheet, and that extra resistor will still fit in your circuit, hopefully. So I'd recommend you to use a pull-up resistor between 10kOhm and 100kOhm (as suggested in the datasheet excerpt) to the input voltage.
Input pins should NEVER be left floating else your circuit will do unpredictable things. Output pin are OK to left unconnected. They will not be "floating" because they are driven to some voltage by them self.
In your case the "interrupt" pin can be left floating as it is an output pin. This pin in an open-collector, which is a output of special kind (that can be connected to other open-collector output) but can be left unconnected.