We may replace $Y$ with an affine open neighbourhood of $y$. Write $Y = \Spec(A)$ and $A = \colim A_i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma \ref{limits-lemma-descend-finite-presentation} we can find an $i$ and a morphism $f_i : X_i \to \Spec(A_i)$ of finite presentation, endowed with a section $s_i : \Spec(A_i) \to X_i$ whose base change to $Y$ recovers $f$ and the section $s$. After possibly increasing $i$ we may also assume that $f_i$ is flat, see Limits, Lemma \ref{limits-lemma-descend-flat-finite-presentation}. Let $y_i$ be the image of $y$ in $Y_i$. Note that $X_y = (X_{i, y_i}) \times_{y_i} y$. Hence $X_{i, y_i}$ is geometrically reduced, see Varieties, Lemma \ref{varieties-lemma-geometrically-reduced-upstairs}. By Lemma \ref{lemma-base-change-connected-along-section} it suffices to prove the lemma for the system $f_i, s_i, y_i \in Y_i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring. \medskip\noindent Assume $Y$ is the spectrum of a Noetherian ring. Since $f$ is of finite presentation, i.e., of finite type, we see that $X$ is a Noetherian scheme too, see Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}. Let $x \in X^0$ be a point lying over $y$. By Topology, Lemma \ref{topology-lemma-constructible-neighbourhood-Noetherian} it suffices to prove that for any irreducible closed $Z \subset X$ passing through $x$ the intersection $X^0 \cap Z$ is dense in $Z$. In particular it suffices to prove that the generic point $x' \in Z$ is in $X^0$. By Properties, Lemma \ref{properties-lemma-locally-Noetherian-specialization-dvr} we can find a discrete valuation ring $R$ and a morphism $\Spec(R) \to X$ which maps the special point to $x$ and the generic point to $x'$. We are going to think of $\Spec(R)$ as a scheme over $Y$ via the composition $\Spec(R) \to X \to Y$. By Lemma \ref{lemma-base-change-connected-along-section} we have that $(X_R)^0$ is the inverse image of $X^0$. By construction we have a second section $t : \Spec(R) \to X_R$ (besides the base change $s_R$ of $s$) of the structure morphism $X_R \to \Spec(R)$ such that $t(\eta_R)$ is a point of $X_R$ which maps to $x'$ and $t(0_R)$ is a point of $X_R$ which maps to $x$. Note that $t(0_R)$ is in $(X_R)^0$ and that $t(\eta_R) \leadsto t(0_R)$. Thus it suffices to prove that this implies that $t(\eta_R) \in (X_R)^0$. Hence it suffices to prove the lemma in the case where $Y$ is the spectrum of a discrete valuation ring and $y$ its closed point. \medskip\noindent Assume $Y$ is the spectrum of a discrete valuation ring and $y$ is its closed point. Our goal is to prove that $X^0$ is a neighbourhood of $X^0_y$. Note that $X^0_y$ is open and closed in $X_y$ as $X_y$ has finitely many irreducible components. Hence the complement $C = X_y \setminus X_y^0$ is closed in $X$. Thus $U = X \setminus C$ is an open neighbourhood of $X^0_y$ and $U^0 = X^0$. Hence it suffices to prove the result for the morphism $U \to Y$. In other words, we may assume that $X_y$ is connected. Suppose that $X$ is disconnected, say $X = X_1 \amalg \ldots \amalg X_n$ is a decomposition into connected components. Then $s(Y)$ is completely contained in one of the $X_i$. Say $s(Y) \subset X_1$. Then $X^0 \subset X_1$. Hence we may replace $X$ by $X_1$ and assume that $X$ is connected. At this point Lemma \ref{lemma-connected-flat-over-dvr} implies that $X_\eta$ is connected, i.e., $X^0 = X$ and we win.