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 $p_k<3p_k<p_l$ 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. 
 A: 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).
A: You can use the simple fact that the numbers $n$, $n+2$, $n+4$ are all incongruent mod $3$, so that at least one of them must be a multiple of $3$. Thus the only triple of numbers $(n,n+2,n+4)$ consisting of all primes is $(3,5,7)$.
If $p_k$ denotes the $k^{th}$ prime and $k$ is the largest index for which $p_k - 2$ is not prime, then $p_{k+2} - 2 = p_{k+1}$ and $p_{k+1} - 2 = p_k$. Thus $(p_k,p_k+2,p_k+4)$ is a triple of primes, implying $p_k = 3$, which is an obvious contradiction.
A: Note that if a prime $p>5$ is of the form $3n+2$ then $p-2$ is not prime. 
There are infinitely many primes of the form $3n+2$.  This follows at once from Dirichlet's Theorem, but it can be shown directly (Pf:  were the list finite we could list all the examples, $\{p_1, \dots, p_k\}$ but then $P=3*\prod {p_i}-1$ is prime to everything on our list and is clearly not the product of primes of the form $3n+1$.)
A: 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:


*

*There are infinite number of primes

*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.
A: Consider multiples of $3$, and add $2$ in them. By Dirichlet's theorem, there are infinitely many primes in $\{ 3n+2\}_n$, and collecting them, subtract $2$ from them, we get non-primes.
A: 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
