Let $F$ be a field. Let $\Gamma$ be an indexing set. Let $\lbrace \theta_i\rbrace_{i\in\Gamma}$ be a collection of of elements of $\overline{F}$. Let $L=F(\theta_i\text{ }|\text{ }i\in\Gamma)$ (the smallest subfield of $\overline{F}$ containing $\lbrace\theta_i\rbrace_{i\in\Gamma}$ and $F$). Let $K\subseteq L$ be an extension of $F$ such that $[K:F]$ is finite.
We know that $K=F(a_1,\dots,a_n)$ for some $a_1,a_2,\dots,a_n\in L$(for example we can take $a_1,\dots,a_n$ to be the basis of $K$ over $F$). But, can we say that $K=F(\theta_{i_1}, \theta_{i_2},\dots,\theta_{i_n})$ for some $i_1,i_2,\dots,i_n\in\Gamma$ ?
In other words, can we say that $K$ is generated over $F$ by a finite subset of $\lbrace \theta_i\rbrace_{i\in\Gamma}$ ?
The answer is no. (Hopefully, I will be more convincing with this answer :-))
Take $F = \mathbb Q$ and $\{\theta_i\,|\,i \in \Gamma\} = \{e^{i \pi/4}\}$. Then $L$ is the 8th cyclotomic field, of degree $\phi(8) = 4$.
Obviously, the only subfields you get by taking subsets of $\{\theta_i\,|\,i \in \Gamma\}$ are $\mathbb Q$ and $L$ itself. But $L$ contains another subfield: $K = \mathbb Q(i)$, of degree two.