Prove that a field $K$ cannot have subfields $k'$ and $k''$ such $k' \cong \mathbb{Q}$ and $k'' \cong \mathbb{F}_{p}$ where $p$ is some prime.
I have tried this by contradiction. Let $K$ a field with subfields $k'$ and $k''$ such $k' \cong \mathbb{Q}$ and $k'' \cong \mathbb{F}_{p}$ for $p$ a prime. Then I consider the prime field of $K$, lets say $k$ is the prime field of $K$. So by a field theory result we have that $k \cong \mathbb{Q}$ or $k \cong \mathbb{F}_{p}$ for some prime $p$ depending if $K$ has characteristic $0$ or $p$ respectively. Without lost of generality lets say that $k \cong Q$ (K has characteristic 0), and $k$ is by definition the smallest of all subfields of $K$, we have that $k \subseteq k'$ and $k \subseteq k''$. But as $k''$ has characteristics $p$ as $k'' \cong \mathbb{F}_{p}$ by hypothesis and $k''$ contains a subfield of order $0$ as $k \subseteq k''$ with $k \cong Q$ we contradict the following fact
The characteristics of a subfield of a field is same as that of the field.
We can obtain an analogous contradiction supposing $k \cong \mathbb{F}_{p}$. Proving that $K$ cannot have subfields $k'$ and $k''$ such $k' \cong \mathbb{Q}$ and $k'' \cong \mathbb{F}_{p}$ where $p$ is some prime.
Is my proof free of any mistakes? Thanks!
Your proof is correct, but extremely roundabout. The prime subfield of $K$ is by definition the intersection of all its subfields; so your proof uses the simple fact that the intersection of two subfields is again a subfield. This simple fact is all you need:
It implies that $k=k'\cap k''$ is a subfield of both $k'=\Bbb{Q}$ and $k''=\Bbb{F}_p$. But both fields have no proper subfields. It follows that $$\Bbb{Q}=k'=k=k''=\Bbb{F}_p,$$ a contradiction.
Or an even simpler proof using the linked question; if such $k'$ and $k''$ exist then $$0=\operatorname{char}k'=\operatorname{char}K=\operatorname{char}k''=p,$$ a contradiction.
Edit: As for the correctness; I just noticed you wrote that
I think you mean characteristic $0$ here.