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As Jim Humphreys has pointed out in the comments, you have to either work over an algebraically closed field, or use scheme language. It is clear that over an algebraically closed field your example does not make any problems. So let us look at your example from the point of view of schemes.

What you are considering is the morphism $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ of $K$-schemes which is obtained as follows: recall that $\mathbb{A}^1_K=\operatorname{Spec} K[x]$, hence to give a morphism as above is the same as to give a homomorphism of $K$-algebras, and our homomorphism is given by $x\mapsto x^2$. Then, the morphism $f$ is not only dominant but surjective as a morphism of schemes.

How does one see this? $\mathbb{A}_K^1$ is the set of prime ideals in $K[x]$. There are two types of these: the prime ideal $(0)$, which defines the generic point, and the maximal ideals. Every maximal ideal is generated by a unique monic irreducible polynomial $f$. Hence if $K$ is not algebraically closed, there are strictly more points in $\mathbb{A}_K^1$ than the generic point and those corresponding to elements of $K$. By this classification of prime ideals it is now an easy exercise to deduce surjectivity of $f$.

However, as you rightly remarked, for $K$ imperfect of characteristic two, the induced map on $K$-rational points is not surjective and the image is "weird" in the sense that it is not of the form $S(K)$ for a subscheme $S\subseteq\mathbb{A}^1_K$. But your argument is wrong. The result from Humphreys that you quote again has to be understood as a statement about schemes, and only in the case of an algebraically closed base field can it be interpreted in the "naive" way as a statement about $K$-rational points.

One correct argument is this: if a subset $S\subseteq\mathbb{A}^1_K$ is constructible (i.e., a subscheme), then it is either finite or the complement of a finite subset. This is because the closed proper subsets of $\mathbb{A}_K^1$ are precisely the finite sets of closed points. But if $K$ is imperfect of characteristic two, than the subset $f(K)\subset K$ is infinite and has infinite complement.

By the way, there is a much easier example: take $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ as above, but with $K=\mathbb{R}$. Then the image of the induced map $\mathbb{R}\to\mathbb{R}$ is the set of nonnegative reals, clearly not "algebraic", constructible", by the same reason.

Finally, here's the correct version of Chevalley's theorem:

Theorem (EGA IV, 1.8.4.) Let $f:X\to Y$ be a finitely presented morphism of schemes (any morphism between varieties over a field is of this type). Then the image of any constructible subset of $X$ under $f$ is a constructible subset of $Y$.

If $K$ is an algebraically closed field and $X$ and $Y$ are varieties over $K$ and $f$ is a morphism of $K$-varieties, then $X(K)$ can be identified with the set of closed (!) points on $X$, and you obtain the "naive" version.

How does one see this? $\mathbb{A}_K^1$ is the set of prime ideals in $K[x]$. There are two types of these: the prime ideal $(0)$, which defines the generic point, and the maximal ideals. Every maximal ideal is generated by a unique monic irreducible polynomial $f$. Hence if $K$ is not algebraically closed, there are strictly more points in $\mathbb{A}_K^1$ than the generic point and those corresponding to elements of $K$. By this classification of prime ideals it is now an easy exercise to deduce surjectivity of $f$.

However, as you rightly remarked, for $K$ imperfect of characteristic two, the induced map on $K$-rational points is not surjective and the image is "weird" in the sense that it is not of the form $S(K)$ for a subscheme $S\subseteq\mathbb{A}^1_K$. But your argument is wrong. The result from Humphreys that you quote again has to be understood as a statement about schemes, and only in the case of an algebraically closed base field can it be interpreted in the "naive" way as a statement about $K$-rational points.

One correct argument is this: if a subset $S\subseteq\mathbb{A}^1_K$ is constructible (i.e., a subscheme), then it is either finite or the complement of a finite subset. This is because the closed proper subsets of $\mathbb{A}_K^1$ are precisely the finite sets of closed points. But if $K$ is imperfect of characteristic two, than the subset $f(K)\subset K$ is infinite and has infinite complement.

By the way, there is a much easier example: take $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ as above, but with $K=\mathbb{R}$. Then the image of the induced map $\mathbb{R}\to\mathbb{R}$ is the set of nonnegative reals, clearly not "algebraic".algebraic", by the same reason.

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As Jim Humphreys has pointed out in the comments, you have to either work over an algebraically closed field, or use scheme language. It is clear that over an algebraically closed field your example does not make any problems. So let us look at your example from the point of view of schemes.

What you are considering is the morphism $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ of $K$-schemes which is obtained as follows: recall that $\mathbb{A}^1_K=\operatorname{Spec} K[x]$, hence to give a morphism as above is the same as to give a homomorphism of $K$-algebras, and our homomorphism is given by $x\mapsto x^2$. Then, the morphism $f$ is not only dominant but surjective as a morphism of schemes. However, as you rightly remarked, for $K$ imperfect of characteristic two, the induced map on $K$-rational points is not surjective and the image is "weird" in the sense that it is not of the form $S(K)$ for a subscheme $S\subseteq\mathbb{A}^1_K$.

By the way, there is a much easier example: take $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ as above, but with $K=\mathbb{R}$. Then the image of the induced map $\mathbb{R}\to\mathbb{R}$ is the set of nonnegative reals, clearly not "algebraic".