Let $R$ be a commutative ring and $P\triangleleft R$ be a prime ideal. Then $Nil(R)\subset P$.
My approach:
Let $r\in R$, $a\in Nil(R)$. Suppose that $r,a\not\in P$, so that $ra\not\in P$ (since $P$ is prime). Since $a\in Nil(R)$, $\exists n\in\mathbb{N}$ such that $a^n=0$, but $0\in P$, so $a^n\in P$. This implies that $(ra)a^{n-1}\in P$. Suppose $ra\not\in P$, then $a^{n-1}\in P$. Then, inductively, $(ra)a\in P$, so $ra\in P$ or $a\in P$, a contradiction. So $r\in P$ or $a\in P$. Also, $a ra\in P$, and $ra = ar\not\in P$, thus $a\in P$.
I think something peculiar is not correct. Would appreciate a review.
You have got the main idea right but your choice of $r$ is superfluous (also I don't understand what your last line of the proof is supposed to show).
Since $a\in \operatorname{Nil}(R)$, there exists some $n$ such that $a^n=0\in P$, as you correctly noted. Why not conclude inductively from this that $a\in P$?