The circuit as shown will work. Remember though that it's a linear regulator, which means that the voltage is dropped by turning excess energy into heat. At 4.5V the drop is small, but if you're dropping to 1.5V (3.5V drop) at 500mA (you shouldn't expect to draw more from USB) then you'll have to deal with 1.75W of heat. The amount of heat that the IC you linked can dissipate depends on the PCB design (did you mean to link a surface mount component?), but in any case 1.75W would be the upper end of what you could expect a TO-220 package component to dissipate. I'd probably use a heat sink or ensure that my load wasn't drawing as much current.
Resistor selection basically doesn't matter. The two aspects that you need to normally consider when selecting resistors are the power rating (1/4W, 1/2W etc) and the tolerance (1%, 5%, 10%). The power rating isn't important in this case (see below) and because you've got a manually adjustable potentiometer the tolerance really isn't important either. Almost any resistor of approximately the right value would do for \$R_2\$.
Both resistors can be low wattage ones. As you've said, the \$I_{ADJ}\$ current is negligible and can be completely ignored. There will also be a current flow through the two resistors from the output voltage to ground, which you can calculate with Ohm's law (\$V=IR\$). In all cases this will be about 7mA.
The capacitors are there for filtering/ripple reduction purposes and their characteristics aren't too important. If they're approximately the same as those suggested in the datasheet (and the voltage rating is above what they'll see) then there shouldn't be any issues.
Your calculations seem to correct. This is just a basic voltage divider calculation. The regulator adjusts the output voltage until it sees 1.24V on ADJ. You can confirm that your calculations are correct with the equation given on wikipedia:
$$V_{div} = V_{in} \times \frac{R_2}{R_1+R_2}$$
Where \$V_{div}\$ is 1.24V, \$V_{in}\$ is your target voltage and you want to solve for \$R_1\$.
$$R1 = \frac{V_{in} \times R_2 - V_{div} \times R_2}{V_{div}}$$
for the 4.5V case therefore:
$$R1 = \frac{4.5 \times 180 - 1.24 \times 180}{1.24} = 473 \Omega.$$
There's inherent variability in electronics which mean that whatever values you calculate won't be exactly right when you wire it up. Wire it up, connect the output to a volt meter and adjust the potentiometer until you have the right voltage.
The 5-second overload rating tells you a couple of things. One, as you note, is an implied "total Joule" rating, which says something about the thermal characteristics of the resistor assembly overall.
The other is that the manufacturer expects that the resistance wire can handle about 700 mA for at least that amount of time. You're asking it to briefly carry twice that current, which may or may not push the resistance wire into a nonlinear region, which could lead to localized failures. With a time constant of 1.65 seconds, it takes about 1.2 seconds for the current to drop from 1.46 A to 700 mA.
I would expect that you'll be fine, but it would be worthwhile to do some tests of your own, running short pulses of current in the range of 1A - 2A through the resistor and monitoring for anomalous resistance changes (i.e., excessive voltage drop).
Or you could just ask the manufacturer directly ...
Best Answer
Then that is fine. The resistor only conducts for a short period of time at start of each AC half cycle when the phototriac is on. And only until there is enough current flowing through it to turn on the main triac. The number determining the dissipation is the turn on current at the triac gate and as it is working fine this is low enough not to cause dissipation problems for the resistor.