Proof that there are infinitely many prime numbers $p$ such that $p-2$ is not prime.

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).

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.

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 $3n+1$.)