Galois over Galois
Here is a full solution of the problem. Since the OP already solved it, I think there is no harm in writing a full solution of it here.
We denote by $Aut(K/k)$ the group of automorphisms of a field extension $K/k$. Let $K/k$ be a not necessarily finite dimensional algebraic extension field. If $K$ is normal and separable over $k$, we say $K/k$ is Galois. If $K/k$ is Galois, we write $G(K/k)$ instead of $Aut(K/k)$.
We need the following characterization of a Galois extension field.
Lemma An algebraic extension field $K/k$ is Galois if and only if the fixed subfield of $K$ by $Aut(K/k)$ is $k$.
Proof. Suppose $K/k$ is Galois. Let $\alpha \in K - k$. Let $f(X)$ be the minimal polynomial of $\alpha$ over $k$. Since $K/k$ is normal and separable, there exists a root $\beta$ of $f(X)$ such that $\alpha \ne \beta$ and $\beta \in K$. Let $\sigma\colon k(\alpha) \rightarrow k(\beta)$ be the unique isomorphism such that $\sigma(\alpha) = \beta$. Since $K/k$ is normal, $\sigma$ can be extended to an automorphism $\sigma'$ of $K/k$. Since $\sigma'(\alpha) = \beta$, we are done.
Conversely suppose the fixed subfield of $K$ by $G = Aut(K/k)$ is $k$. Let $\alpha$ be an element of $K$. Let $f(X)$ be the minimal polynomial of $\alpha$ over $k$. Since $\sigma(\alpha)$ is a root of $f(X)$ for every $\sigma \in G$, the set $S = \{\sigma(\alpha)\mid \sigma \in G\}$ is finite. Let $\sigma_1, \cdots, \sigma_m$ be elements of $G$ such that $\sigma_1(\alpha), \cdots, \sigma_m(\alpha)$ are pairwise distinct and $S = \{\sigma_1(\alpha), \cdots, \sigma_m(\alpha) \}$. Let $g(X) = (X - \sigma_1(\alpha))\cdots (X - \sigma_m(\alpha))$. Since every coefficient of $g(X)$ is fixed by $G$, $g(X) \in k[X]$. Since $g(\alpha) = 0$, $g(X)$ is divisible by $f(X)$. Since each $\sigma_i(\alpha)$ is a root of $f(X)$, $g(X) = f(X)$. Hence $\alpha$ is separable over $k$ and $K/k$ is normal. This completes the proof of the lemma.
Now let $F/K, E/K$ be as in the problem. By the lemma, it suffices to prove that the fixed subfield of $F$ by $Aut(F/K)$ is $K$. Suppose $\alpha \in F$ is fixed by $Aut(F/K)$. Since $G(F/E) \subset Aut(F/K)$, $\alpha$ is fixed by $G(F/E)$. Hence $\alpha \in E$ by the lemma. Since every element of $G(E/K)$ is extended to an element of $Aut(F/K)$, $\alpha$ is fixed $G(E/K)$. Hence $\alpha \in K$ by the lemma. This completes the proof.
I think the above answer assumes that $F/K$ is algebraic and although most of the time when we consider galois extensions we assume that they are algebraic as well, this problem can be solved without assuming so and in a shorter way.
Let $u\in F-K$, we have to prove that there is a $K$-automorphism $\sigma:F\to F$ such that $\sigma(u)\neq u$. Lets consider two cases:
$u\in E$. Since $E/K$ is galois there is a $K$-automorphism $\theta:E\to E$ such that $\theta(u)\neq u$, we extend this $\theta$ to a $K$-automorphism $\sigma:F\to F$ and we are done.
$u\notin E$. Since $F/E$ is galois there is a $E$-automorphism $\sigma:F\to F$ such that $\sigma(u)\neq u$ and this is also a $K$-automorphism since $K\subset E$