I am trying to solve the following problem but to no avail.
Let $F \subsetneq K \subseteq E$ be fields and let $\alpha \in E$ and $F(\alpha)=E$, then $[E:K] < \infty$
If $\alpha$ is algebraic over $F$ then the problem is trivial, I am struggling to prove it for the case it is not as I have never seen such extensions.
As I recall, if $\alpha$ is transcendental over $F$ this result is non-trivial, and goes by the name of Lüroth's theorem, which asserts that, given $\alpha$ transcendental over $F$, any intermediate extension $L$ between $F$ and $F(\alpha)$ is itself purely transcendental over $F$, hence of the form $F(\theta)$ for some $\theta\in L$ transcendental over $F$.
At this point, by taking $L=K$, we conclude that $F(\alpha)$ is algebraic over $K$, by the fact that transcendence bases of $E$ over $F$ have all the same cardinality, and hence its degree over $K$ is finite, for $F(\alpha)$ is a simple extension of $K$.
Edit: You do not really need Lüroth's theorem. Assume $\alpha$ transcendental over $F$.
Then, any element $\beta\in F(\alpha)\setminus F$ is indeed transcendental over $F$: otherwise, setting $\beta=f(\alpha)/g(\alpha)$ for some $f,g\in F[\alpha]$, there would be $c_i$'s in $F$ such that $\sum_{i\leq n} c_if^i(\alpha)/g^i(\alpha)=0$, which would give $\sum_{i\leq n} c_if^i(\alpha)g^{n-i}(\alpha)=0$, which is an algebraic relation for $\alpha$.
Now, take $F\subsetneq K\subsetneq F(\alpha)$: since $K\setminus F$ is non-empty, $K$ has transcendence degree greater than $0$ over $F$. Moreover, its transcendence degree is exactly one: otherwise, $K$ would contain two algebraically independent elements over $F$ and so would $F(\alpha)$, which is false.
Therefore, any $\beta\in K\setminus F$ is a transcendence basis for $F(\alpha)$ over $F$: in particular, $\alpha$ is algebraic over $K$, being algebraic over $F(\beta)$. But $F(\alpha)=K(\alpha)$, and $[K(\alpha):K]$ is finite.