Just need verification of this proof by contradiction:
Let $p_k$ be the largest prime number such that $p_k-2$ is not a prime number. Let $p_l$ be some prime number such that $p_l > 3p_k$. We know that $p_l$ exists because there are infinitely many prime numbers as proven elsewhere.
$p_l-2$ is a prime number. $p_l-4$ is a prime number as well. By induction, all numbers $p_k < x < p_l | x \mod 2 = 1 $ are prime numbers. But $p_k<3p_k<p_l$ and $3p_k \mod 2 = 1$ and $3|3p_k$ so there is a contradiction.
Also would love to see more concise proofs if at all possible.
Edit:
Great that I got so many people trying to provide their own proof but the question was primarily asked so that I could have my proof verified. Only skyking addressed my question, though I do not yet understand his criticism.
Best Answer
It is very simple to construct such infinite sequence:
$$35+60n,37+60n$$
$35+60n$ will generate an infinite amount of composite numbers, all of which are divisible by $5$ (in fact, it will generate only composite numbers).
$37+60n$ will generate an infinite amount of prime numbers, since $37$ and $60$ are coprime integers (according to Dirichlet's theorem on arithmetic progressions).