I have a fairly simply proof.
For it, I keep a running product, starting with 4. I then look at the number just before the running product, in this case 3.
Now 3 is 4*0 + 3, and it happens to be prime, so I multiply my product by 3 so that the next number in my series isn't divisible by 3.
So now my product is 12, and the next number of the form 4k+3 is 11. Also prime, and I multiply it into the product.
Now, for each number in the series, I know it isn't divisible by any of the primes I found before. That leaves 2 possibilities.
1) The number is prime.
2) The number is equal to a NEW prime of the form 4k+3 multiplied by a number of the form 4k+1. (Modulo arithmetic)
Either way, I can examine the number to arrive at a new 4k+3 prime, which then gets multiplied into the product for one more prime of the form 4k+3.
As I can do this indefinitely, there must therefore be an infinite number of 4k+3 primes.
Pardon the ugliness of the proof, but it works as far as I can see.
(It doesn't, however, work for 4k+1 primes, as a 4k+1 value can be the product of 2 4k+3 values..)