Basically I had to proves that there are infinite many primes congruent to $3$ modulo $4$, and the professor said "look at Euclid's proof for infinitude of primes"
So basically I did the following. Assume otherwise and assume there are finite many. Define $S={p_0,p_1,...,p_k}$With $p_0=3$. Then consider $x=4p_1...p_k+3.$ Since $x$ is congruent to $3$ modulo $4$ it suffices to show it would be a prime number, contradicting the fact that $S$ contained all such numbers. Note that no member of $S$ divides $x$, and that $2$ does not divide $x$. Note that $x$ is not a product of primes congruent to $1$ modulo $4$, because the product of such primes gives a number congruent to $1$ modulo $4$. Thus, $x$ is not divisible by any primes, so it should be prime. Contradiction because clearly $x\notin S$.
Now my problem is that I did not get credit for this proof because of the following: I wrote the sentence "$x$ is a prime", and they grader argued that $x$ is rarely going to be a prime, but my point is that if $S$ indeed contained all the primes, then we could construct another prime.
I did not get the points back, but I just want to know whether my proof is faulty. If so, why?