Is a separable extension of an extension field also a separable extension of the base/ground field? A field is called separably closed if the only separable algebraic extension is the trivial one. A separable closure of a field $K$ is a separable algebraic extension $K ⊆ K^{\text{sep}}$ with $K^{\text{sep}}$ separably closed.
I want to show that a separable closure $K^{\text{sep}}$ of $K$ is algebraically closed iff $K$ is perfect (i.e. every algebraic extension of $K$ is separable).
I have already proven

I) Every field has a separable closure.
II) Every pair of separable closures of $K$ is $K$-isomorphic.
III) For every tower $K ⊆ L ⊆ M$ of algebraic extensions we have $$ K⊆ L \text{ and } L ⊆ M  \text{ are both separable } \iff K ⊆ M \text{ is separable.}$$

So here we go. Suppose a separable closure $K^{\text{sep}}$ of $K$ is algebraically closed. Let $L/K$ be an arbitrary algebraic extension. By (I) $L$ has a separable closure $L^{\text{sep}}$. Now if $L^{\text{sep}}$ were a separable extension of $K$ as well, then by (II) $K^{\text{sep}}$ and $L^{\text{sep}}$ would be $K$-isomorphic, so we would obtain a tower
$$K ⊆ L ⊆ L^{\text{sep}} \cong_K K^{\text{sep}}. 
$$
of algebraic extensions. As $K ⊆ K^{\text{sep}}$ is separable by definition, (III) implies that $K ⊆ L$ must be too, so $K$ is perfect.
However
All we really know is that $L ⊆ L^{\text{sep}}$ is separable. i.e. for all $x ∈ L^{\text{sep}}$ the minimal polynomial $f^x_L ∈ L[X]$ has no multiple roots in $\bar{L}$. This unfortunately does not imply that $f^x_K ∈ K[X]$ has no multiple roots in $\bar{K}$, as $f^x_L$ divides $f^x_K$, not the other way around. Is the reasoning so far correct? If so, how do I show the final ingredient?
 A: Let's do the direction you attempted. I think your difficulty is that you are confused about the following point: generally one fixes a particular algebraic closure of the ground field, and then an extension of that ground field is taken to mean an extension in that particular choice of algebraic closure. This just ensures that things like the compositum always "make sense". One can do this because every algebraic extension (living anywhere) admits an inclusion into any particular algebraic closure; in particular all algebraic closures are (non-canonically) isomorphic.
Let $L/K$ algebraic. So either we take $L$ to already be inside of $\bar K = K^{sep}$ (under your assumption) or there is an inclusion of $L$ into $\bar K = K^{sep}$. Either way, $L$ is separable (in the second case it is isomorphic to a separable extension, which is the same thing). So all algebraic extensions are separable, which means $K$ is perfect.
Conversely, if $K$ is perfect then all of its algebraic extensions are separable, so in particular the algebraic closure is separable!
