Let $k =\mathbb F_p$, the finite field of $p$ elements and $F = k(t, u)$ where $t$ and $u$ are independent variables. Let $α$ and $β$ be roots of $x^p − t$ and $x^p − u$ respectively. Let $L = F(α, β)$. Consider the intermediate fields $F ⊂ F(α+λβ) ⊂ L$ as $λ$ varies over all elements of $F$. I want to show there are infinitely many intermediate fields between $F$ and $L$ in the following way.
So let's say $λ ≠ μ$ are two elements of $F$ such that $F(α + λβ) = F(α + μβ)$, how to show that $α, β ∈ F(α + λβ)$? If I can do this then I can conclude that $F(α + λβ) = F(α, β)$ which is not possible right? :O
Assume that $F(\alpha + \lambda\beta) = F(\alpha + \mu\beta)$ for some $\lambda, \mu \in F$. Then we have that $\alpha + \mu\beta \in F(\alpha + \lambda\beta)$ and so:
$$(\mu - \lambda)\beta = \alpha + \mu\beta - (\alpha + \lambda\beta) \in F(\alpha + \lambda\beta)$$
This implies that $\beta \in F(\alpha + \lambda\beta)$, as $(\mu - \lambda)$ is an element of $F$. From here it's not hard to conclude that $\alpha = \alpha + \lambda\beta - \lambda\beta \in F(\alpha + \lambda\beta)$. Therefore as you've noted we get that $F(\alpha + \lambda\beta) = F(\alpha, \beta)$.
To see that this equality is impossible, note that:
$$(\alpha + \lambda\beta)^p = \alpha^p + \lambda^p\beta^p = t + \lambda^pu \in F$$
This means that $[F(\alpha + \lambda\beta):F] = p$, while $[F(\alpha,\beta):F]=p^2$, which is a contadiction.