It is a theorem that if $M[G]$ is a generic extension $M$, then for every model $N$ of ZFC with $M \subset N \subset M[G]$, $\ N$ is some generic extension of $M$ (and is, in fact, $M[G\cap D]$ for some complete subalgebra $D$ of the complete algebra $B$ over which $G$ is $M$-generic).
This made me wonder, is there a real $r$ and a countable transitive model $M$ such that $r$ is not in any forcing extension of $M$?
On
While not as clever as $0^\#$ or adding ordinals, here is something a bit odd.
If $M$ is a countable transitive model of $\sf ZFC$, then it has a class forcing which adds a real $r$ such that $M[r]\models V=L[r]$. Moreover this can be minimal, so any set in $M[r]$ is either in $M$ or it codes $r$.
Such a real is not set generic over $M$. So as far as set forcing is concerned, it is not in any forcing extension of $M$.
Yes - many reals, such as $0^\#$, cannot be added by forcing to a model in which they do not already exist.
Without going that far, for any countable model $M$ the real coding $M$ is not addable by forcing. (Or, similarly, any real coding a well-ordering of length $\ge Ord(M)$.)