This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.
For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.
Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.
Proof.
We show that if $k \subseteq K \subseteq k(t)$ is a sequence of field
extensions such that $\operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K \simeq k(X)$ for a transcendental element $X$ over $k$. Since
$\operatorname{trdeg}_k K = 1$, it is enough to find some $a \in K$ such that
$K = k(a)$.
First, the second extension $K \subseteq k(t)$ must be algebraic, so
$t$ is algebraic over $K$. Let
$$
f(X) = X^n + a_1X^{n-1} + \cdots + a_n \in K[X]
$$
be the minimal polynomial of $t$ over $K$. Since $t$ is
transcendental over $k$, we cannot have all $a_i \in k$, and so there is some
$i$ such that $a_i \in K \smallsetminus k$. We will show that in this case, $K
= k(a_i)$. We know that the $a_i \in K \subseteq k(t)$,
and so we may write
$$
a_i = \frac{u(t)}{v(t)},
$$
where $u,v \in k[t]$ are relatively prime, and where at least one of them of
positive degree by the assumption that $a_i \notin k$. Now consider the following polynomial:
$$
F(X) = u(X) - a_iv(X) \in k(a_i)[X] \subseteq K[X].
$$
Since $F(t) = 0$, we have that $f(X) \mid F(X)$ in $K[X]$ by minimality of
$f(X)$, and so
\begin{equation}
u(X) - a_iv(X) = f(X)g(X).\tag{1}\label{eq:0105star}
\end{equation}
where $g \in K[X]$. We then claim the following:
Claim. $g \in K$.
Showing the Claim would conclude the proof, for then we have a sequence of
extensions
$$
k \hookrightarrow k(a_i) \hookrightarrow K \hookrightarrow k(t)
$$
where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence
$[k(t) : K] \ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.
To prove the Claim, the first step is to get rid of all denominators: by
multiplying \eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we
get a relation
$$
c(t)\bigl(u(X)v(t) - v(X)u(t)\bigr) = f_1(t,X)g_1(t,X),
$$
where $c(t) \in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) \in k[t,X]$ are obtained from $f$ resp. $g$ by
multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$.
Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$
to get a relation
\begin{equation}
u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),\tag{2}\label{eq:0105rel3}
\end{equation}
where now $f_2(t,X),g_2(t,X) \in k[t,X]$ are obtained from $f,g$ by
multiplication by a nonzero element in $k(t)$. The trick is now to look at the
degrees in $t$ on both sides. First,
$$
\deg_t\bigl(u(X)v(t) - v(X)u(t)\bigr) \le \max\bigl\{\deg u(t),\deg
v(t)\bigr\}.
$$
Letting $f_2(t,X) = \gamma_0(t)X^n + \cdots + \gamma_n(t)$, we see that
$$
\frac{\gamma_i(t)}{\gamma_0(t)} = a_i(t) = \frac{u(t)}{v(t)},
$$
where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact,
$$
\deg_t\bigl(f_2(t,X)\bigr) \ge \max\bigl\{\deg u(t),\deg v(t)\bigr\}.
$$
By looking at degrees in $t$ on both sides of the relation
\eqref{eq:0105rel3}, we have that $\deg_t(g_2(t,X)) = 0$, hence $g_2 \in k[X]$.
Now we show that $g_2 \in k$. For sake of contradiction, suppose that
$g_2 \notin k$. Then, there is a root $\gamma \in \overline{k}$ such that
$g_2(\gamma) = 0$, which implies that $u(\gamma)v(t) = v(\gamma)u(t)$ by \eqref{eq:0105rel3}. But
since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we
must have $u(\gamma) = 0 = v(\gamma)$. This contradicts that $u,v$ are
relatively prime, and so $g_2 \in k$.
Finally, since $g \in K[X]$ and $g = g_2 \cdot (\text{element of
$k(t)$})$, we have that $g \in K$. $\blacksquare$
