Suppose that $F$ is a field of characteristic $p > 0$ and $[E:F] = p$. Then suppose that there is some $\alpha \in E\setminus F$ with an inseparable minimal polynomial over $F$.
Then every $\alpha \in E\setminus F$ has an inseparable minimal polynomial over $F$.
How does one see this?
Suppose $\alpha \in E\setminus F$ has an inseparable minimal polynomial $f_\alpha \in F[x]$.
Then deg$(f_\alpha) | p$ and thus is $= p$ (else it must have degree $1$ but that contradicts $\alpha \not \in F$).
But now we may use the fact that $f_\alpha(x) = g(x^{p^d})$ for some irreducible separable $g$ and some $d \geq 1$. But since degree of $f_\alpha$ is $p$, we thus have that $d = 1$ and $g(x) = x-\alpha$.
I.e, $f_p(x) = (x-\alpha)^p = x^p - \alpha^p$ and so $\alpha^p \in F$.
Moreover, one can also conclude I think that $E = F(\alpha)$ (as $F(\alpha) \leq E$ but $[F(\alpha) : F] = [E:F] = p$)
So now we have that $E = F(\alpha)$ and $\alpha^p \in F$.
This I think results in that if $\beta \in E = F(\alpha)$ then $\beta^p \in F$ as well- as any $\beta \in F(\alpha)$ is like $\beta = \sum_{i \leq p}a_i \alpha^i$ where the $a_i \in F$ and then $\beta^p = \sum_{i \leq p}a_i^p\alpha^{p^i} \in F$. And so $x^p - \beta^p = (x-\beta)^p$ would be the minimal (and inseparable) polynomial for $\beta$- this is minimal since $(x-\beta)^p$ is irreducible in $F$- if $(x-\beta)^k \in F[x]$ for any $k \leq p$ then $(x-\beta)^k = x^k - k\beta x^{k-1} + \cdots \in F[x]$ which means that $k\beta in F[x]$, but then since $\beta \not \in F$ this can only occur if $k = p$ and so $(x-\beta)^p$ is irreducible in $F[x]$ after all. Thus $(x-\beta)^p$ is the minimal and inseparable polynomial for every $\beta \in E\setminus F$ And thus every $\beta$ is inseparable, proving the conclusion.
Does that seem right? Is there a simpler way to do this?
Let $K$ be the set of all elements of $E$ which are separable over $F$. This is a field and hence $K$ is an intermediate field between $E$ and $F$. Thus, $[E:K][K:F] = [E:F] = p$, whence it follows that either $[K:F] = 1$, in which case every element of $E \setminus F$ is inseparable or $[E:K] = 1$, in which case every element of $E$ is separable over $F$.
I think if you unpack all of the facts I stated without proof, you'll get back your proof, but hopefully this helps clarify something.