9V battery and resistor are connected in series - the voltage across the resistor is < 9V. Why?

They are simulating a (more or less) real 9V battery. They've modeled the battery as an ideal voltage source of 9V with a series resistance of ~1.5 ohms.

^{simulate this circuit – Schematic created using CircuitLab}

If you work this out, the current is 9V/(11.5 ohms) = 0.783A, so the voltage across the 10 ohm resistor must be 7.83V.

The 7.83 volts tells you precisely what the internal series resistance of the battery is. Open circuit it is 9 volts but under load it drops to 7.83 volts - the current thru the 10 ohm is clearly 783 mA. This current also flows thru the internal resistance of the battery to lose 9 minus 7.83 volts (1.18 volts).

1.18 volts lost at 783 mA means the internal resistance is 1.507 ohms.

All the above is about ohms law and applying it.