Let $k \subseteq k(\alpha)$ be a simple extension with $\alpha$ transcendental. Show that if $E$ is an intermediate field that properly contains $k$, then $E \subseteq k(\alpha)$ is a finite extension.
(Attempted) proof: Let $\beta \in E - k$. We will consider the extension $k(\beta) \subseteq k(\alpha)$; if this is finite, then so is $E \subseteq k(\alpha)$.
Because $\alpha$ is transcendental, $k(\alpha)$ is isomorphic to the rational expressions in $\alpha$ over $k$. This implies that $\beta = f(\alpha)/g(\alpha)$, where $f$ and $g$ are polynomials, and $g$ non-zero. Because $\beta \notin k$, the degree of $f$ and $g$ are not both $0$. This lets us construct a polynomial in $k(\beta)[x]$ that $\alpha$ is a root of: namely, $\beta g(x) - f(x)$. Because $\alpha$ is algebraic over $k(\beta)$, $k(\beta) \subseteq k(\alpha)$ is finite.
Does this look about right? I am particularly concerned about bolded clause.