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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