The remark mentioned in the title states that if $L/K$ is a Galois extension, an $E$ is an intermediate field such that $E/K$ is a Galois extension, then any $K$-automorphism of $L$ restricts to a $K$-automorphism of $E$.
This should be a consequence of Theorem 4(i), section 3.5, which says that an algebraic extension $G/F$ is normal iff any $F$-homomorphism of $G$ into an algebraic closure $\bar G$ of $G$ restricts to a $F$-automorphism of $G$.
In particular since $E/K$ is Galois, so normal, Theorem $4$ implies that any $K$-automorphism of an algebraic closure $\bar E$ restricts to a $K$-automorphism of $E$. However, how do I use the fact that $L/K$ (and so $L/E$) is Galois, to prove that such implication holds also with $L$ in place of $\bar E$? I don't think $L$ is necessarily an algebraic closure of $E$.
Embed $L$ into an algebraic closure $\overline{L}$. Note that since $L$ is algebraic over $E$, $\overline{L}$ is also an algebraic closure of $E$.
Now let $\sigma$ be a $K$-automorphism of $L$. Since $L\subseteq \overline{L}$, we can view $\sigma$ as a $K$-homomorphism $L\to \overline{L}$. Then $\sigma|_E$ is a $K$-homomorphism $E\to\overline{L}$. Since $E$ is normal over $K$, the image of $\sigma|_E$ is $E$, so $\sigma|_E$ is a $K$-automorphism of $E$, as desired.