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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