Choose a diagram $$ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X_V \ar[r] \ar[d] & X \ar[d]^f \\ Y' \ar[r]^g & V \ar[r] & Y } $$ as in Lemma \ref{lemma-make-components-generic-fibre-geometrically-irreducible}. Note that the generic fibre of $f'$ is geometrically connected (for example by Lemma \ref{lemma-base-change-fibres-nr-geometrically-connected-components}). Suppose that the lemma holds for the morphism $f'$. This means that there exists a nonempty open $W \subset Y'$ such that every fibre of $X' \to Y'$ over $W$ is geometrically connected. Then, as $g$ is an open morphism by Morphisms, Lemma \ref{morphisms-lemma-etale-open} all the fibres of $f$ at points of the nonempty open $V = g(W)$ are geometrically connected, see Lemma \ref{lemma-base-change-fibres-nr-geometrically-connected-components}. In this way we see that we may assume that the irreducible components of the generic fibre $X_\eta$ are geometrically irreducible. \medskip\noindent Let $Y'$ be the reduction of $Y$, and set $X' = Y' \times_Y X$. Then it suffices to prove the lemma for the morphism $X' \to Y'$ (for example by Lemma \ref{lemma-base-change-fibres-nr-geometrically-connected-components} once again). Since the generic fibre of $X' \to Y'$ is the same as the generic fibre of $X \to Y$ we see that we may assume that $Y$ is irreducible and reduced (i.e., integral, see Properties, Lemma \ref{properties-lemma-characterize-integral}) and that the irreducible components of the generic fibre $X_\eta$ are geometrically irreducible. \medskip\noindent At this point suppose that $X_\eta = X_{1, \eta} \bigcup \ldots \bigcup X_{n, \eta}$ is the decomposition of the generic fibre into (geometrically) irreducible components. Let $X_i$ be the closure of $X_{i, \eta}$ in $X$. After shrinking $Y$ we may assume that $X = \bigcup X_i$, see Lemma \ref{lemma-cover-generic-fibre-neighbourhood}. Let $Z_{i, j} = X_i \cap X_j$. Let $$ \{1, \ldots, n\} \times \{1, \ldots, n\} = I \amalg J $$ where $(i, j) \in I$ if $Z_{i, j, \eta} = \emptyset$ and $(i, j) \in J$ if $Z_{i, j, \eta} \not = \emptyset$. After shrinking $Y$ we may assume that $Z_{i, j} = \emptyset$ for all $(i, j) \in I$, see Lemma \ref{lemma-empty-generic-fibre}. After shrinking $Y$ we obtain that $X_{i, y}$ is geometrically irreducible for each $i$ and all $y \in Y$, see Lemma \ref{lemma-geom-irreducible-generic-fibre}. After shrinking $Y$ some more we achieve the situation where each $Z_{i, j} \to Y$ is flat and of finite presentation for all $(i, j) \in J$, see Morphisms, Proposition \ref{morphisms-proposition-generic-flatness}. This means that $f(Z_{i, j}) \subset Y$ is open, see Morphisms, Lemma \ref{morphisms-lemma-fppf-open}. We claim that $$ V  = \bigcap\nolimits_{(i, j) \in J} f(Z_{i, j}) $$ works, i.e., that $X_y$ is geometrically connected for each $y \in V$. Namely, the fact that $X_\eta$ is connected implies that the equivalence relation generated by the pairs in $J$ has only one equivalence class. Now if $y \in V$ and $K \supset \kappa(y)$ is a separably closed extension, then the irreducible components of $(X_y)_K$ are the fibres $(X_{i, y})_K$. Moreover, we see by construction and $y \in V$ that $(X_{i, y})_K$ meets $(X_{j, y})_K$ if and only if $(i, j) \in J$. Hence the remark on equivalence classes shows that $(X_y)_K$ is connected and we win.