Suppose first that $K=F(\alpha)$ Since $K/F$ is finite, $\alpha$ is algebraic over $F$ Let $m(x)$ be the minimal polynomial of $\alpha$ over $F$ Now, let $L$ be an intermediary field with $F\subseteq L\subseteq K$ and let $m'(x)$ be the minimal polynomial of $\alpha$ over $L$ Also, let $L'$ be the field generated by the coefficients of the polynomial $m'(x)$ Thus, the minimal polynomial of $\alpha$ over $L'$ is still $m'(x)$ and $L'\subseteq L$ By the properties of the minimal polynomial, and since $m(\alpha)=0$ we have a divisibility $m'(x)|m(x)$ and so: $$[K:L]=\deg(m'(x))=[K:L'].$$ Since we know that $L'\subseteq L$ this implies that $L'=L$ Thus, this shows that each intermediary subfield $F\subseteq L \subseteq K$ corresponds with the field of definition of a (monic) factor of $f(x)$ Since the polynomial $f(x)$ has only finitely many monic factors, we conclude that there can be only finitely many subfields of $K$ containing $F$

Now suppose conversely that there are only finitely many such intermediary fields $L$ If $F$ is a finite field, then so is $K$ and we have an explicit description of all such possibilities; all such extensions are generated by a single element. So assume $F$ (and therefore $K$ are infinite. Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be a basis for $K$ over $F$ Then $K=F(\alpha_1, \ldots, \alpha_n)$ So if we can show that any field extension generated by two elements is also generated by one element, we will be done: simply apply the result to the last two elements $\alpha_{j-1}$ and $\alpha_j$ repeatedly until only one is left.

So assume $K=F(\beta,\gamma)$ Consider the set of elements $\{\beta+a\gamma\}$ for $a\in F^{\times}$ By assumption, this set is infinite, but there are only finitely many fields intermediate between $K$ and $F$ so two values must generate the same extension $L$ of $F$ say $\beta+a\gamma$ and $\beta+b\gamma$ This field $L$ contains $$ \frac{(\beta+a\gamma)-(\beta+b\gamma)}{a-b} = \gamma $$ and $$ \frac{(\beta+a\gamma)/a-(\beta+b\gamma)/b}{1/a-1/b} = \beta $$ and so letting $\alpha = \beta+a\gamma$ we see that $$ F(\alpha)=L=F(\beta,\gamma)=K. $$