Just invoking Bertrand's postulate might be enough for the OP, but I
also expect something towards actually starting to address the
Goldbach conjecture, which of the existing answers at the time I
posted this bounty, there was only one.
Okay. Consider then even number $2p + 2$. Where $p$ is prime so that there are no primes between $p$ and $2p$.
This is a counter example to the gold conjecture.
Let $2p + 2 = q + r$ where $q$ and $r$ are prime. Wolog assume $q \ge r$.
If $q > p$ then $q \ge 2p+1$ so $r \le 1$. That can't work. If $q \le p$ then $q + r \le 2p < 2p+2$.
So $2p + 2$ can not be written as sum of two primes.
So the goldbach conjecture would be disproven if we could find such a prime $p$. But Bertrand's postulate which has beed proven directly says there is no such $p$.
So the goldbach conjecture has not been disproven.
If the number $2p + 2 = q + r$ where $p,q,r; q \ge r$ are primes it would have to be that $p < q < 2p$ but as there will always be some prime between $p$ and $2p$, this is not a problem.
If the goldbach conjecture is true, a (maybe) interesting conesequence is that for any prime $p$, there is a prime $q$ so that $p < q < 2p$ and $2p-q+2$ is also prime.
Ex. $2 < 3 < 2*2; 2*2+2-3 = 3$. $3< 5< 6$ and $8-5 = 3$. $5 < 7 < 10; 12-7 = 5$. and ... $13 < 17,\langle19\rangle,23 < 26$ and $28 -17,\langle19\rangle,23 = 11,\langle9\rangle, 5$ etc.