Here is a proof. Let $t$, $K$ and $F$ be as stated.

**Preliminary reduction:** $t$ is defined over $F$.

**Proof:** Let $t(x) = p(x)/q(x)$, with $p=\sum_{i=0}^r p_i x^i$ and $q=\sum_{j=0}^s q_j x^j$. For every $x \in F$, let $y = t(x)$; then $\sum q_j x^j y - \sum p_i x^i$ is a linear constraint on the $p_i$ and $q_j$, with coefficients in $F$. Since there is a nonzero solution to these constraints in $K$, there is also a solution in $F$; call it $(p'_0, \ldots, p'_r, q'_0, \ldots, q'_s)$. So we have a rational function $p'/q'$ such that $p'(x)/q'(x) = p(x)/q(x)$ for all $x \in F$. Since $F$ is infinite, this means that $p/q = p'/q'$. QED

**Theorem:** Let $F$ be infinite. Let $t \in F(x)$ be nonconstant and separable. Let $L$ be the subfield of $F(x)$ generated by $t(g(x))$, as $g$ ranges over all nonconstant rational functions in $F(x)$. Then $L=F(x)$.

**Proof:** Since $F(t) \subseteq L \subseteq F(x)$, we see that $F(x)/L$ is a separable finite extension. By Luroth's theorem, $L = F(u)$ for some $u \in F(x)$, and this $u$ is a separable map $\mathbb{P}^1_F \to \mathbb{P}^1_F$. Suppose for the sake of contradiction that $u$ has degree $d>1$.

Since $F$ is infinite, we can find two points in $\mathbb{P}^1_F$ where $t$ takes two different values; using an automorphism of $\mathbb{P}^1_F$, we may assume that $t(\infty) \neq t(0)$.

Since $u$ is separable, for all but finitely many $y \in \mathbb{P}^1_F$, the preimage $u^{-1}(y)$ in $F^{\mathrm{alg}}$ has size $d$. Since $F$ is infinite, we can choose $\zeta \in \mathbb{P}^1_F$ so that $u^{-1}(u(\zeta))$ has size $d$ (with the preimage taken in $F^{\mathrm{alg}}$.)

Let $\alpha$ be an element of $u^{-1}(u(\zeta))$ other than $\zeta$. So $\alpha$ lies in some finite extension of $F$ (possibly $F$ itself). Let $p$ be the minimal polynomial of $\alpha$ and set $g(x) = p(x)/(x-\zeta)$. So $(t \circ g)(\alpha) = t(0) \neq t(\infty) = (t \circ g)(\zeta)$. So $t \circ g$ takes different values on $\alpha$ and on $\zeta$, and thus cannot be in $F(u)$. We have found an element of $L$ which is not in $F(u)$, a contradiction. The theorem is proved. QED

By the Theorem, $x$ can be written as $h(t \circ g_1, t \circ g_2, \ldots, t \circ g_N)$ for some rational functions $g_i \in F(x)$ and some $h \in F(y_1, \ldots, y_N)$. So, for any $x \in K$, we have
$$x = h(t(g_1(x)), \ldots, t(g_N(x)) ).$$

Now, all of the $g_i(x)$ are in $K$. So, by hypothesis, all the $t(g_i(x))$ are in $F$. So $h(t(g_1(x)), \ldots, t(g_N(x)) )$ is in $F$. In short, we have shown that, for every $x \in K$, we have $x \in F$, which is what we wanted.