TITLE: Is there a prime $p$ whose successor is greater than $2p$?
QUESTION [5 upvotes]: Toying with Goldbach's Conjecture, I encountered myself in a situation where the following question arose.
Is there a prime $p$ whose successor is greater than $2p$?
You see. If the answer to this question is true, then Goldbach's Conjecture's disproven.

REPLY [5 votes]: As many of the other answers have noted, there is no prime $p$ such that its successor is greater than $2p$. This is a trivial consequence of Bertrand's postulate which states that

For any natural number $n>1$, there exists some prime $p$ such that $$n<p<2n$$

This ensures that for all primes $p_n$, its successor, $p_{n+1}$, must fall within the range $p_n<p_{n+1}<2p_n$. If you do not understand why that is, think about $p_n$ as a natural number. By Bertrand's postulate there must exist a prime greater than $p_n$ but less than $2p_n$. This garuntees the existence of a prime $p_{n+k}$ such that $p_n<p_{n+k}<2p_n$.   As such, $p_{n+1} \leq p_{n+k}$ and hence must lie in this range as well. 
The proof of Bertrand's postulate is quite long. If you wish to see one, I would recommend Erdös' elementary proof. Particularly, the following paper by Michael Tang presents the proof in a fashion particularly easy to follow if you have little background in number theory. http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNy8yLzcyZTgyOGRhZDgxMmQ5MTY0ODIwOTJjZTUyZWQ0OWI4ZjIzYWVmLnBkZg==&rn=QmVydHJhbmQucGRm 
Although I do not  believe that this was your original intention, I will now address the link between this problem and the Goldbach Conjecture. You are correct when you say that the existence of a prime $p$ whose successor is larger than $2p$ would disprove the Goldbach conjecture. In fact, there exists a deeper link between Bertrand's postulate and the strong Goldbach conjecture. The Goldbach conjecture implies Bertrand's postulate. A proof of this can be found here. https://proofwiki.org/wiki/Goldbach_implies_Bertrand 
However, although Goldbach's conjecture implies Bertrand's postulate, this does not work the other way around. That is, Bertrand's postulate does not imply the Goldbach conjecture (otherwise the Goldbach conjecture would be proven).