This isn't a rigorous mathematical proof, but more a logical narrowing-down. All primes are odd except for 2. No primes end in the number 5, except for 5, because anything else that ends in 5 is divisible by 5 and hence not prime. Any set of primes where each successive prime was 2 more than the last one would have to have last digits from among these, in order: 1, 3, 5, 7, 9. There is no way of doing that without including 5 as a last digit. Hence the only way to do it is {3, 5, 7}.
Nevermind, that doesn't work. You could still have a run of primes where the last digits went 7, 9, 1 or 9, 1, 3. I may still have generally been on the right track, though; perhaps there is some way of showing that if the last digits went 7, 9, 1 or 9, 1, 3 that one of those would have to be divisible by something?
Actually I think I have just worked it out. If you take any number, it is either a multiple of 3 or you can add either 2 or 4 to make it a multiple of 3. So it's impossible to have three numbers generated by adding 2 and 4 to a given number and not have at least one of them be divisible by 3. Again, I don't know how to write that up as a proper proof.
Why the heck am I studying biochemistry and not maths again? This stuff is so much fun!