If we consider the prime field $\mathbb{Q}$, and the field extension $\mathbb{Q}\subseteq\mathbb{C}$, then for any natural $n$ we can find $a_1,...,a_n\in\mathbb{C}$ algebraically independent over $\mathbb{Q}$. To prove this, we use induction on $n$, where the case $n=1$ is trivially true. Now if there are $a_1,...,a_{n-1}$ algebraically independent, we can (if I'm not mistaken) consider an algebraic closure of $\mathbb{Q}(a_1,...,a_{n-1})$, which would still be in $\mathbb{C}$, but also still countable, since $\mathbb{Q}$ and $\mathbb{Q}(a_1,...,a_{n-1})$ are. Hence there must be an $a_n$ as desired.
Is this proof correct so far?
I'm also interested in a more general statement about fields in arbitrary characteristic, something like:
Let $F$ be a prime field. Then for any natural $n$ we can find a field $K$ containing $F$ and elements $a_1,...,a_n\in K$ algebraically independent over $F$.
Unfortunately, I'm not very used to fields, and finite fields especially. I don't know if this statement is true anyway. Could you tell me? If it should be true, is there a way (in finite characteristic) to choose one $K$ for all $n$, and even maybe use a similar way of proving the statement?