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 $p_l > 3p_k$. 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

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.

Answer

First of all some criticism, the proof seem to be somewhat overly complicated. But I see no fault in it.

It's not that clear that $p_l-4$ is a prime number, not using the fact that $p_k$ being the largest prime such that $p_k-2$ is not a prime at least.

The statement however follows quite easily from two observations:

1. There are infinite number of primes


2. There are infinite number of odd composite numbers (non-primes)

For each odd composite number $k$ there exists a smallest prime $p(k)>k$. Now since it's the smallest such prime you have that $p(k)-2$ is not a prime ($p(k)-2\ge k$, either it's $k$ and composite or between $k$ and $p(k)$ and therefore composite).

The first observation is a direct consequence of that $3(2n+1)$ is composite. The second is because otherwise (if there were finitely many primes) the product of all primes plus 1 would be another prime.