**Proof.**
Part (1) is immediate from Lemma 38.4.5 and the fact that the inverse image of a standard open under a morphism of affine schemes is a standard open, see Algebra, Lemma 10.17.4.

Let $W \subset Y$ as in (2). Because $Y \to S$ is smooth it is open, see Morphisms, Lemma 29.34.10. Hence we can find a standard open neighbourhood $S'$ of $s$ contained in the image of $W$. Then the fibres of $W_{S'} \to S'$ are nonempty open subschemes of the fibres of $Y \to S$ over $S'$ and hence geometrically irreducible too. Setting $Y' = W_{S'}$ and $Z' = \pi ^{-1}(Y')$ we see that $Z' \subset Z$ is a standard open neighbourhood of $z$. Let $\overline{h} \in \Gamma (Z, \mathcal{O}_ Z)$ be a function such that $Z' = D(\overline{h})$. As $i : Z \to X$ is a closed immersion, we can find a function $h \in \Gamma (X, \mathcal{O}_ X)$ such that $i^\sharp (h) = \overline{h}$. Take $X' = D(h) \subset X$. In this way we obtain a standard shrinking as in (2).

Let $U \subset X$ be as in (3). We may after shrinking $U$ assume that $U$ is a standard open. By More on Morphisms, Lemma 37.47.4 there exists a standard open $W \subset Y$ neighbourhood of $y$ such that $\pi ^{-1}(W) \subset i^{-1}(U)$. Apply (2) to get a standard shrinking $X', S', Z', Y'$ with $Y' = W_{S'}$. Since $Z' \subset \pi ^{-1}(W) \subset i^{-1}(U)$ we may replace $X'$ by $X' \cap U$ (still a standard open as $U$ is also standard open) without violating any of the conditions defining a standard shrinking. Hence we win.
$\square$

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