It is not always true that $L^{\operatorname{Aut}(L/k)} \subset E$ (counterexample given at the bottom), but it is true in your scenario.
I also feel that the application of the Galois correspondence you mentioned is happening in the case where $[E:k] < \infty$. Perhaps there are some more checks to be done in the case where $[E:k] = \infty$, but I am not sure.
Here is how I would prove the result you're after:
Let $E/k$ be a finite extension. Let $G = \operatorname{Aut}(E/k)$ and let $k_0 = E^G$. Let $I$ be the compositum of all subfields $F$ of $E$ such that $F \supset k$ and $F$ is purely inseparable over $k$. Note that $I$ is purely inseparable over $k$, and in fact, $I$ consists of all elements $\alpha$ in $E$ that are purely inseparable over $k$.
First, we show that $I \subset E^G$.
Let $\alpha \in I$ and let $\sigma \in G$. We want to show that $\sigma(\alpha) = \alpha$. Since $\alpha$ is purely inseparable over $k$, $\operatorname{Irr}(\alpha,k,X)$ has only one distinct root, namely $\alpha$. Since $\sigma(\alpha)$ is also a root of $\operatorname{Irr}(\alpha,k,X)$, it must be that $\sigma(\alpha) = \alpha$. Thus, $\alpha \in E^G$.
Second, we remark that $E/E^G$ is separable.
This is proved in Proposition 6.11 of Chapter V in Lang's Algebra (pages 251-252, third edition), so we skip the proof here.
Third, we introduce the hypothesis that $E^G \subset I$.
But, we proved that $I \subset E^G$, so we have that $E^G = I$. Hence, we have a chain of extensions $E \supset E^G \supset k$ such that $E^G/k$ is purely inseparable and $E/E^G$ is separable. Hence, by Proposition A in this note by Joseph Lipman, there exists a separable extension $K/E$ such that $K/k$ is normal. Moreover, if $H = \operatorname{Aut}(K/k)$, then $K^H = I$.
To show that $E/k$ is normal, it suffices to show that $E/I$ is normal.
This is because $E/I$ normal and $I/k$ purely inseparable together imply that $E/k$ is normal (see here or here).
Now, since $E/E^G = E/I$ is separable, showing that $E/I$ is normal is equivalent to showing that $E/I$ is Galois. Since $E/k$ is a finite extension, this is equivalent to showing that the fixed field of $\operatorname{Aut}(E/I)$ equals $I$.
To do this, we will show that $\operatorname{Aut}(E/I)$ equals $G = \operatorname{Aut}(E/k)$. Then, the fixed field of $\operatorname{Aut}(E/I)$ will be $E^G = I$, proving the result.
Since $\operatorname{Aut}(E/k) \supset \operatorname{Aut}(E/I)$, it suffices to show that $\operatorname{Aut}(E/k) \subset \operatorname{Aut}(E/I)$. So, let $\sigma \in G$. We need to show that $\sigma \in \operatorname{Aut}(E/I)$, that is, that $\sigma$ fixes every element of $I$. Extend $\sigma$ to an embedding of $K$ in an algebraic closure of $K$, also denoted $\sigma$. Since $K/k$ is normal, $\sigma$ is an automorphism of $K$ over $k$. Hence, $\sigma \in H = \operatorname{Aut}(K/k)$ and we have shown that $K^H = I$. Thus, $\sigma$ fixes every element of $I$, as was to be shown.
Notice that we have used the facts used in Lipman's note to prove that the fixed field of $\operatorname{Aut}(K/k)$ is a subfield of $E$, which is where you were stuck. In the same note, Lipman gives an example of an extension of fields for which this does not happen: let $\Bbb{F}_2$ be the field of two elements, let $Y$, $Z$ be indeterminates, let $k=\Bbb{F}_2(Y, Z)$, and let $E=k(x)$, $x$ being a root of $$f(X) = X^4 + YX^2 + Z = 0.$$ Then it turns out that there is no normal extension $K$ of $k$ containing $E$ such that $K^{\operatorname{Aut}(K/k)} \subset E$, based on the facts proved in the note.
