The exercise is to prove this proposition using the following lemmas:
1) Prove Lemma A: every prime ≠ 2 is either of the form $4n + 1$ or $4n + 3$
2) Prove Lemma B: the product of numbers of the form $4n + 1$ is of the form $4n + 1$.
I'll omit the proofs of the lemmas, since I'm not doubtful about their demonstrations. Again I ask if there is any flaw on the following proof.
Suppose, by contradiction, that there are finitely many primes of the form $4n + 3$, say, $p_1, p_2, ..., p_k.$ Define $N = 4p_1...p_k - 1 = 4(p_1...p_k - 1) + 3.$ $N$ > $p_k$ and is of the form $4n + 3$, therefore is not prime. So $N$ must be divisible by some prime. Since $N$ is not divisible by any $p_i$ or 2, this prime divisor must be > $p_k$ and consequently, due to lemma A, of the form $4n + 1$. Now, the fundamental theorem of arithmetic states that any non-prime number must be factorable in prime numbers. Since $N$ is not divisible by any $p_i$, there must be primes $f_1, ..., f_l$ on which $N$ is factorable. From the hypothesis and from lemma A follows that these primes must be of the form $4n + 1$ and, by lemma B, so must be $N$, a contradiction.