**Proof.**
Let $X'$ be a nonsingular projective model of $X$. Then $X'$ and $X$ have isomorphic nonempty open subschemes. In particular $X'$ is geometrically irreducible as $X$ is (some details omitted). Thus we may assume that $X$ is projective.

Assume $X$ is proper. Then $X_ K$ is proper and hence the normalization $(X_ K)^\nu $ is proper as a scheme finite over a proper scheme (Varieties, Lemma 33.27.1 and Morphisms, Lemmas 29.44.11 and 29.41.4). On the other hand, $X_ K$ is irreducible as $X$ is geometrically irreducible. Hence $X_ K^\nu $ is proper, normal, irreducible, and birational to $(X_ K)_{red}$. This proves (1) because a proper curve is projective (Varieties, Lemma 33.43.4).

Proof of (2). As $X$ is proper and we have (1), we can apply Varieties, Lemma 33.27.4 to find $K/k$ finite purely inseparable such that $Y_ K$ is geometrically normal. Then $Y_ K$ is geometrically regular as normal and regular are the same for curves (Properties, Lemma 28.12.6). Then $Y$ is a smooth variety by Varieties, Lemma 33.12.6.

If $Y_ K$ is geometrically reduced, then $Y_ K$ is geometrically integral (Varieties, Lemma 33.9.2) and we see that $H^0(Y_ K, \mathcal{O}_{Y_ K}) = K$ by Varieties, Lemma 33.26.2. This proves (3) because a smooth variety is geometrically reduced (even geometrically regular, see Varieties, Lemma 33.12.6).

If $Y_ K$ is smooth, then for every extension $\Omega /K$ the base change $(Y_ K)_\Omega $ is smooth over $\Omega $ (Morphisms, Lemma 29.34.5). Hence it is clear that $Y_\Omega = (Y_ K)_\Omega $. This proves (4).
$\square$

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