I know I have already asked a question regarding this proof. However, I wanted to see if my reformulation of this proof (with my better understanding in my own words and after some time) is correct.
Prove: There are infinitely many primes with remainder $2$ when divided by $3$:
Proof: Suppose not. Suppose there are finitely many primes with remainder $2$ when divided by $3$. That is:
There are finitely many primes of the form $p = 3k+2$ for some $k\in \mathbb{Z}$. Thus, we can construct a list of these odd primes (to exclude 2) and take their product:
$N = p_1p_2\cdots p_r$ for some $r \in \mathbb{Z}$
Now consider $3N+2$:
We know that $3N+2 \in \mathbb{Z}$ and has a factorization into primes. We also know that since $3N+2$ is of the form $3\cdot(\text{some integer}) + 2$ there is at least one prime factor of $3N+2$ of the form $3\cdot(\text{some integer}) + 2$. We also know that none of the primes $p_1p_2,\ldots,p_r$ divide $3N+2$ because if they did we would have for some prime $p_i$ in our list:
$p_i\mid(3N+2) \land p_i\mid N \implies p_i\mid 2$ Which cannot happen since our list of primes excludes $2$. Thus we have the two contradicting statements:
There exists a prime factor of the form $3k+2$ and there is no prime factor (from our finite list $p_1,p_2,\ldots,p_r$) of the form $3k+2$. Therefore, the supposition is false and there are infinitely many primes with remainder $2$ when divided by $3$.
Is this correct?