Why does a closed circuit with a pull-up resistor output "0"?
The switch is essentially an ideal conductor, with a negligible resistance that can be ignored. When closed, the node between the pull up resistor and the switch is pulled to ground or 0 potential. The resistor, as ohm laws would dictate, carries the 5v across it, "using" the voltage. Since the path of least resistance is a direct connecting to ground, the node is pulled to ground. 

When the switch is open, the node is open, and no current flows, bringing the node to 5V through the weak resistor. 

Think of the switch as a direct wire.