The Stacks project

Proof. Write $S = \mathop{\mathrm{Spec}}(R)$ and let $\pi \in R$ be a uniformizer. To get a contradiction assume that $X_\eta $ is disconnected. This means there exists a nontrivial idempotent $e \in \Gamma (X_\eta , \mathcal{O}_{X_\eta })$. Let $U = \mathop{\mathrm{Spec}}(A)$ be any affine open in $X$. Note that $\pi $ is a nonzerodivisor on $A$ as $A$ is flat over $R$, see More on Algebra, Lemma 15.22.9 for example. Then $e|_{U_\eta }$ corresponds to an element $e \in A[1/\pi ]$. Let $z \in A$ be an element such that $e = z/\pi ^ n$ with $n \geq 0$ minimal. Note that $z^2 = \pi ^ nz$. This means that $z \bmod \pi A$ is nilpotent if $n > 0$. By assumption $A/\pi A$ is reduced, and hence minimality of $n$ implies $n = 0$. Thus we conclude that $e \in A$! In other words $e \in \Gamma (X, \mathcal{O}_ X)$. As $X$ is connected it follows that $e$ is a trivial idempotent which is a contradiction. $\square$