The voltage across a capacitor discharging into a fixed resistance decays exponentially. The time constant is RC, where C is the capacitance, and R is the resistance between the terminals of the resistor.
$$
V(t)=V(0)e^{\frac{-t}{RC}} \\
$$
You're right that the capacitor starts at 20V, because that's the voltage across the 200 ohm resistor after the capacitor is charged. So you know V(0), and you know C. All you're missing is R. To analyze that properly, think of the switch as a resistor of 0 ohms when closed, and don't worry about the current source. The parallel combination of a 200 ohm resistor and a 0 ohm resistor is 0 ohms. The series combination of a 50 ohm resistor and a 0 ohm resistor is 50 ohms. So the 50 ohm resistor is the only one that matters when determining your discharge constant.
The current through the 200 ohm resistor depends on the voltage across the 200 ohm resistor. The voltage across the 200 ohm resistor is the same as the voltage across the 0 ohm resistor (the closed switch). V=IR, so what's the voltage across that pair of resistors when the switch is closed?
And yes, the voltage across an ideal resistor can change instantaneously. Keep in mind though, there's no such thing as an ideal resistor in the physical world. Everything has capacitance to everything else.
Without the resistor, if you press a button, you'll short 5V to 0V and BAD THINGS will happen.
On another hand, if we leave bad things for a moment as a thought experiment, what level would arduino read?
5V ----+---- 0V
The voltage at the + would be defined by the ratio of resistances of wires going from 5V and 0V to a + point. This would be tricky to set up properly, so it's better to put a big resistor (much bigger than wire resistance) and have the result predictable.
Edit
by BAD THINGS I mean - what does happen when you short a battery? In theory unlimited current flows from positive terminal to negative, in reality this current is limited by internal resistance of the battery, battery heats up, potentially explodes or leaks. If you are powering from USB - usb circuitry inside computer fries (this rarely happens as computers usually have protection circuitry). If you are powering circuit from a wall-wart adapter, a fuse in it might blow or it can catch a fire if it's of some extremely crappy chinese origin. If your power supply is strong and robust enough (like lead acid battery or lithium cell), a power trace on a PCB can burn off. In this particular case nothing bad would happen - arduino has a resettable PTC fuse on board, it will just reboot.
Best Answer
When resistors overheat, a variety of things can happen, some of which will increase resistance and some of which will decrease resistance--typically by creating alternate current paths. Overheated resistors are not particularly likely to ever represent a "dead short", but high-value resistors may have their resistance drop by multiple orders of magnitude. Note that subjecting a resistor to sustained moderate overloading may yield very different behavior from subjecting it to severe overvoltage conditions. If an overvoltage condition causes an arc within a resistor's insulating material, the resistance may be greatly reduced as long as the arc is sustained, and the susceptibility to arcing may be permanently reduced, even if the resistance under low-voltage conditions is relatively unaffected.