The Stacks project

Proof. Choose an étale morphism $U \to X$ where $U$ is an affine scheme. Then $U$ is of finite type over $R$. Let $u \in U$ be in the special fibre. The local ring $A = \mathcal{O}_{U, u}$ is essentially of finite type over $R$, hence Noetherian. Let $\pi \in R$ be a uniformizer. Since $X$ is flat over $R$, we see that $\pi \in \mathfrak m_ A$ is a nonzerodivisor on $A$ and since the special fibre of $X$ is reduced, we have that $A/\pi A$ is reduced. If $a \in A$, $a \not= 0$ then there exists an $n \geq 0$ and an element $a' \in A$ such that $a = \pi ^ n a'$ and $a' \not\in \pi A$. This follows from Krull intersection theorem (Algebra, Lemma 10.51.4). If $a$ is nilpotent, so is $a'$, because $\pi$ is a nonzerodivisor. But $a'$ maps to a nonzero element of the reduced ring $A/\pi A$ so this is impossible. Hence $A$ is reduced. It follows that there exists an open neighbourhood of $u$ in $U$ which is reduced (small detail omitted; use that $U$ is Noetherian). Thus we can find an étale morphism $U \to X$ with $U$ a reduced scheme, such that every point of the special fibre of $X$ is in the image. Since $X$ is proper over $R$ it follows that $U \to X$ is surjective. Hence $X$ is reduced. Since the generic fibre of $U \to \mathop{\mathrm{Spec}}(R)$ is reduced as well (on affine pieces it is computed by taking localizations), we conclude the same thing is true for the generic fibre. $\square$