Why does this method of proof for $\overline{\mathbb{F}_{p}}$ not apply for $\mathbb{F}_{p}$ Let $\overline{\mathbb{F}_{p}}$ be the algebraic closure of the finite field $\mathbb{F}_{p}$. Consider $F:= \overline{\mathbb{F}_{p}}(x,y)$ and let $K:=\overline{\mathbb{F}_{p}}(x^{p},y^{p})$.
Below I will show that $\overline{\mathbb{F}_{p}}(x,y)/\overline{\mathbb{F}_{p}}(x^{p},y^{p})$ is not a simple extension by exhibiting infinitely many intermediate subfields. My concern is, I was told that the same proof does not work if we simply consider $\mathbb{F}_{p}$ and not its algebraic closure. However, I have come across proofs of this result that proceed the same way. So I would like to ask is: in what way does the method of proof need to be changed (if at all)? Since I have seen a handful of proofs that look almost identical, there must be something subtle that's different. 
PROOF: We can show that $\overline{\mathbb{F}_{p}}(x,y)/\overline{\mathbb{F}_{p}}(x^{p},y^{p})$ is not a simple extension by using the result that $K/F$ (finite) extension is simple iff there exist finitely many subfields of $K$ containing $F$. This is done by noting that $[F : K] = p^{2}$ and that the subfields $F(x + cy)$ for $c \in\overline{\mathbb{F}_{p}}$ are all of degree $p$ over $\overline{\mathbb{F}_{p}}(x^{p}, y^{p})$, noting that 
$$(x+cy)^{p} = x^{p} + c^{p}y^{p} \in \overline{\mathbb{F}_{p}}(x^{p}, y^{p})$$
Then we can conclude by showing that if $F(x+c_{1}y) = F(x+c_{2}y)$ for $c_{1} \neq c_{2}$ then we would have that $\overline{\mathbb{F}_{p}}(x,y) = F(x+cy)$ which is impossible by degree considerations. Hence there are infinitely many such subfields and so the extension cannot be simple.
 A: I don't know what "they" told you but the extension $E=\mathbb{F}_{p}(x,y)/L=\mathbb{F}_{p}(x^{p},y^{p})$ is also not simple!     
Indeed $[E:L]=p^2$, whereas any rational function $\phi(x,y)=\frac{P(x,y}{Q(x,y)}\in E$ (with $P(x,y), Q(x,y)\neq0\in \mathbb{F}_{p}[x,y]$) satisfies $\phi(x,y)^p=\frac{P(x^p,y^p)}{Q(x^p,y^p)}\in \mathbb{F}_{p}(x^{p},y^{p})=L$, so that the degree of $\phi(x,y)$ over $L$ is at most $p$ and we thus cannot have  $L(\phi)=E$.
In other words no element  $\phi \in \mathbb{F}_{p}(x,y)=E$ generates the extension $E/L$, which is thus not simple.
Remarks
a) You had better forget about this business of finitely many intermediate fields, which is  a red herring for the problem at hand.
b) Did you notice that I used $(P(x,y))^p=P(x^p,y^p)$ ? Do you believe that ? Why ?  
Edit
Since mistertanaka asks for an infinite family of intermediate fields $  L=\mathbb F_p(x^p,y^p) \subset M \subset E=\mathbb F_p(x,y)$, here is such a family.
  To  $\phi\in L=\mathbb F_p(x^p,y^p)$ associate the field $M_\phi=L(x+\phi\cdot y)$.
I claim that the $M_\phi$ are different for different $\phi$ which will immediately show the existence of an  infinite family of intermediate fields $M_\phi$.
So,  suppose $\phi\neq \psi$ but $M_{\phi}=M_{\psi}=:M$.
 Then $M$ has degree $\leq p$ and $x+\phi\cdot y,x+\psi\cdot y\in M$ so that $(\phi-\psi)\cdot y\in M$ and thus $y\in M$.
But this implies $x\in M$ too so that $M=\mathbb F_p(x,y)=E$, which is absurd because $[M:L]\leq p$, whereas $[E:L]=p^2$.
A: There are only finitely many choices for $c \in \mathbf{F}_p$.
Let $E/F$ be any field extension in characteristic $p$. If we take the compositums $E' = \overline{\mathbf{F}}_p E$ and $F' = \overline{\mathbf{F}}_p F$, I assert that $E/F$ is simple if and only if $E'/F'$ is simple.
