Yeah, pretty much.
One minor observation... according to datasheet, the segments can do 25mA each but if you'd run them at such high current they'd be super bright. You'd want to run them at 5-10mA tops, this will be bright enough and the on-off effect will make them seem brighter anyway.
Some pics can source a maximum of about 15 mA per pin or per whole chip, I don't remember exactly which one was now, so you should use npn transistors to turn on the
You have the voltage drop on the diode, 2.1v min , 2.5v max so just assume 2.3v ... then you have ohm's laws v=IxR ... so you can figure out how to limit the current to what you want.. for example if you use 5v and want 10mA (0.01A) on segments , you need to drop 5v (input) - 2.3 (vdrop on segment) = 2.7v on the resistor so 2.7v = 0.01 x R therefore R = 2.7 / 0.01 = 270mA
But keep in mind there's some voltage drop on the transistors at both anode and cathode, so you'll have to take out those drops from the formula. Basically, 330ohm is reasonable for 5-10mA on each segment.
You could use a ULN2003A or ULN2803A, these are DIP chips that have 7 (first) or 8 (the second) Darlington transistors rated for 500mA each so a bit overkill but makes it easy to connect the seven segment display, because these particular numbers have a built in resistor at the base of each darlington transistor (note that there are similar chips like ULN2801 or 2803 without A at the end which don't have the base resistors or like ULN2804/A which is the cmos version and needs at least 6v to work)
The ULN2803A has a 2.7kOhm base resistor so it only needs 1/10 of mA or so from the microcontroller to turn the segment on and as a bonus, you can run the leds at something like 12v while your microcontroller sends 5v at the base.
So with such dip IC you only need the ic and resistors at each output, sized so as to keep in consideration the 1-1.2v drop on the darlington.