Suppose that $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$, and let $F \leq \overline{\mathbb{Q}}(x)$ be a subfield. Then I want to show that either $F/\mathbb{Q}$ or $\overline{\mathbb{Q}}(x)/F$ is an algebraic extension.
I guess I get the idea somewhat- If $F/\mathbb{Q}$ is not algebraic then that must mean that in some form $F = K(x)$ where $K \leq \mathbb{Q}$, and then obviously we do get that $\overline{\mathbb{Q}}(x)/K(x)$ is algebraic. That is the intuitive answer, but how does one formalize this?
If $F/\mathbb{Q}$ is not algebraic, then there is $y \in F \subseteq \overline{\mathbb{Q}}(x)$ that is transcendental. Does this directly imply that $x \in F$? I'm sort of confused here. And if we do get that $x \in F$, we get that $F \geq \mathbb{Q}(x)$, but can one say that $F = K(x)$ for some $K \geq \mathbb{Q}$?
Your idea that if $F$ is not algebraic over $\newcommand{\Q}{\mathbb{Q}}$ then it must be of the form $F = K(x)$ is not right, but something very similar does work.
If $F$ is not algebraic over $\Q$, then we must have one element in $F$ which is transcendental over $\Q$. Let's call this element $t \in \newcommand{\QQ}{\overline{\Q}}\QQ(x)$. Set $K = \Q(t)$. Clearly, $\Q \leq K \leq F \leq \QQ(x)$ where $A \leq B$ represents $B$ is an extension field of $A$.
It is clear that if $\QQ(x)$ is algebraic over $K$, then it is algebraic over $F$, so without loss of generality, we assume that $F = \Q(t)$ for some element $t \in \QQ(x)$.
Now, what does it mean for $t$ to be an element of $\QQ(x)$? One has that $$t = \frac{p(x)}{q(x)},$$ where $p,q \in \QQ[x]$. Can you use this create a non-zero polynomial with coefficients in $F = \Q(t)$ such that $x$ is a root of it?
In other words, can you describe a polynomial $f(X) \in F[X] = \Q(t)[X]$ such that $f(x) = 0$ and $f \not\equiv 0$?