Proof that there are infinitely many prime numbers $p$ such that $p-2$ is not prime. 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 $p_l &gt; 3p_k$. 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 &lt; x &lt; p_l | x \mod 2 = 1 $ are prime numbers. But $p_k&lt;3p_k&lt;p_l$ 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.
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).
We can have the obvious case where the last digit of p is 7. By drichlet's theorem on arthmetic progressions, we can generate a sequence of 5*n* + 7, where there are infinitely many primes with last digit as 7

Note: I considered the obvious case since (10k+7)-2 is divisible by 5, thus not prime