If $F, K$ are fields, $F$ algebraically closed, and $F subseteq K$ then $K = F$?

I want to show that an algebraically closed field $F$ cannot be contained in a larger field $K$. So if $F subseteq K$ then $F = K$ for all fields $K$.
Here’s my attempt at a proof:

For contradiction, Say $F subsetneq K$. Hence there is an element $k in K$, $k not in F$.

• if $k$ is algebraic over $F$, then we adjoin $k$ into $F$. So consider $F(k) equiv F[X] / (p)$ for some polynomial $p in F[X]$. Since $F(k)$ should be a field, we need $(p)$ to be maximal [ring quotient maximal ideal is field]. Since we have that $F$ is algebraically closed, irreducible polynomials [which generate maximal ideals] are going to be linear. Hence $(p)$ is maximal and F$(k)$ is a field iff $p = (x – f_star)$ for some $f_star in F$. But $F[X]/ (x- f_star) simeq F$, since quotienting by $(x – f_star)$ keeps only degree 0 polynomials, which are the constant elements. So we have that $F(k) = F$. Hence we cannot have $k not in F$, giving us the desired contradiction.
• If $k$ is transcendental over $F$, I feel that some argument ought to hold, but I don’t know how to proceed.

Is there a counter-example, where $F$ is algebraically closed, while still possessing an extension $K = F(k)$ for some $k$ that is transcendental over $F$? The only algebraically closed field I have experience with, $mathbb C$, does not allow such a thing to happen as far as I am aware.