Does an inseparable extension have a purely inseparable element?

The standard counterexample is to consider $F = \mathbf{F}_p(x,y)$ and $K = F(\alpha)$ where $\alpha$ is a root of $f = T^{p^2} + x T^p + y \in F[T]$ (where $f$ is irreducible by Gauss' Lemma). Clearly $K/F$ is non-separable of degree $p^2$ and $F' := F(\alpha^p)$ is a degree-$p$ subextension that is separable over $F$, so this is the maximal separable subextension. Hence, $[K:F]_s = p$ and so $[K:F]_i = p$. To rule out the existence of an element of $K-F$ that is purely inseparable over $F$ it suffices to show that $F'$ is the unique degree-$p$ subextension of $K/F$.

Suppose $E$ is another such subextension, so it must be purely inseparable over $F$ (as otherwise it would be separable and hence the composite $F'E$ would be separable yet has to exhaust $K$ for $F$-degree reasons). By multiplicativity of separable degree in towers, it follows that $[K:E]_s=p$, so the degree-$p$ extension $K/E$ is separable. But $K=E(\alpha)$, so the degree-$p$ minimal polynomial $g \in E[T]$ of $\alpha$ is separable. In a splitting field $E'/E$ of $f$ over $E$, $f$ is a product of $p$th powers of $p$ monic linear factors and $g$ is the product of just these monic linear factors (since $g|f$ in $E[T]$ and $g$ is separable over $E$ of degree $p$).

Hence, in $E'[T]$ we must have $g^p=f$. But this latter equality then must hold in $E[T]$ as well. It then follows that $g = T^p + uT + v$ for some $u, v \in E$ satisfying $u^p = x, v^p = y$. But the extension $\mathbf{F}_p(x^{1/p}, y^{1/p}) \supset F$ has degree $p^2$ and so cannot lie inside the degree-$p$ extension $E$ of $F$. This is a contradiction, so no such $E \ne F'$ exists inside $K/F$.

I think this is one of the counterexamples mentioned by grghxy on MO : take $F=\mathbb{F}_2(x,y)$ and consider in $F[t]$ the polynomial $f(t)=t^4+xt^2+y$. Using Gauss' Lemma for example it is easy to see that $f(t)$ is irreducible in $F[t]$, and then we can take the field extension $K= F(\alpha)=F[t]/(f(t))$. This is a degree $4$ extension and it is inseparable, since $f'(t)=0$. Now we want to show that no element $\beta \in K$ is purely inseparable over $F$. Suppose that $\beta \in K$ is purely inseparable over $F$: since $[K:F]=4$ it must be either $\beta^2\in F$ or $\beta^4\in F$. We can write $$ \beta = a_0 + a_1\alpha +a_2\alpha^2 + a_3\alpha^3 \qquad a_i \in F $$ and then after some computations (I hope they are correct) one can see that $$ \beta^2 = [a_0^2+a_2^2+a_3^2xy] + [a_1^2+a_2^2x+a_3^2x^2+a_3^2y]\alpha^2 $$ so that $\beta^2\in F$ if and only if $$ a_1^2+a_2^2x+a_3^2x^2+a_3^2y=0 $$ Multiplying this equation by the denominators of $a_i^2$, we can suppose that $a_i \in \mathbb{F}_2[x,y]$, but then we observe that $a_3^2y$ has odd degree in $y$, whereas the other terms have even degree in $y$. This implies that $a_3=0$, so that we are left with $$ a_1^2+a_2^2x=0 $$ but again the term $a_2^2x$ has odd degree in $x$ and the other one has even degree, so that we are left with $a_1=a_2=0$. This proves that if $\beta^2 \in F$ then $\beta\in F$.

Now, suppose that $\beta^4 \in F$, then from what we have proves above it must be that $\beta^2\in F$, so that again $\beta\in F$.