Does every curve over a number field have infinitely many rational functions of fixed degree Let $X$ be a curve over a number field $K$ of genus $g\geq 2$. Does there exist an integer $d$ such that $X$ has infinitely many rational functions (i.e., finite morphisms $f:X\to \mathbf{P}^1_K$) of degree $d$? 
If yes, can we choose/bound $d$ in terms of $K$ and $g$?
Note. This is not the same question as in the title. The answer to the question in the title is in fact negative as the example below shows for $d=2$.
Note: I want rational functions to be really different. So we mod out by the action of the automorphism group of $\mathbf{P}^1_K$. That is, $f$ and $\sigma\circ f$ are the same if $\sigma$ is an automorphism of $\mathbf{P}^1_K$.
Example 1. We can not have $d=2$. Hyperelliptic maps are unique.
Question 2. How does the answer to this question change if we replace $K$ by an algebraically closed  field $k$?
Example 2. Let $X$ be a general curve of odd genus  $g\geq 3$ over an algebraically closed field. Then, it has an infinite number of (really different) gonal morphisms. In fact, this family is one-dimensional.
Idea. I think the question can be reduced to a question on $\mathbf{P}^1_K$. In fact, it suffices to show that there are infinitely many (really different) rational functions on $\mathbf{P}^1_K$ of some degree, say $3$. In fact, once you know this, composing some  morphism $f:X\to \mathbf{P}^1_K$ with such a rational function gives infinitely many rational functions of degree $d \leq 3 \deg f$. The only problem is then finding (in a controlled way) a morphism $f:X\to \mathbf{P}^1_K$. This is a hard problem, but let's allow finite base change if necessary...
 A: I think you are looking for rational subextensions of $K(X)$ of index $d$. 
First in the field $K(t)$ of rational functions of $\mathbb P^1_K$, there are infinitely many subextensions of index $2$: Consider the $K(t^2+\lambda t+1)$ for $\lambda\in K$. 
If $t^2+\lambda t + 1\in K(t^2+\mu t+1)$, then $(\lambda - \mu)t\in K(t^2+\mu t+1)$, hence $\lambda - \mu=0$ because otherwise $t\in K(t^2+\mu t +1)$. Therefore we get infinitely many subextension of index $2$ (this has no much to do with number fields, just need $K$ is infinite). 
Now apply your idea and you get a positive answer to your question. I don't understand what you mean by a hard problem. Any inclusion $K(x)\subset K(X)$ corresponds to a unique morphism $X\to \mathbb P^1_K$ of degree $[K(X) : K(x)]$. 
For a bound on $d$: first there always exists a morphism $X\to \mathbb P^1_K$ of degree $\le 2g-2$ (use the canonical morphism, recall $g\ge 2$). So a rough upper bound is $2(2g-2)$.
Edit When are there infinitely many gonal morphisms (i.e. morphisms to $\mathbb P^1$ of smallest degree), including the case $g=1$ ? Let $f : X\to \mathbb P^1_K$ be given by an $f\in L(D)$ for some effective divisor $D$ of degree $d$ with $\deg f=d$. Suppose $D$ is base point free and if $\dim L(D)\ge 3$, then there are infinitely many gonal morphisms if $K$ is infinite. 
Proof. Write $D=\sum_{1\le i\le n} a_i[x_i]$ with $a_i>0$. As $D$ is base point free, $L(D)\ne L(D-[x_i])$. So there exists $f_1\in L(D)\setminus \cup_i L(D-[x_i])$ because $K$ is inifnite. Let $f_2\in L(D) \setminus (K+Kf_1)$ (recall that $\dim L(D)\ge 3$). 
Let's show that 
$$K(f_2)\ne K(f_1).$$ 
Otherwise $f_2=(\alpha f_1+ \beta)/(\gamma f_2+ \delta)$ with $\alpha, \beta, \gamma, \delta\in K$ and $\gamma\ne 0$. So we can write 
$$af_2+b=1/(\gamma f_1+c), \quad a, b, c\in K, \ a\ne 0.$$ 
So $1/(\gamma f_1+c)\in L(D)$. As 
$$(\gamma f_1+c)_{\infty}=(f_1)_{\infty}=D,$$
this is impossible. Let it is easy to see that $K(f_1+\lambda f_2)$, when $\lambda$ runs through $K$, form an infinite family of pairwise distinct rational subextension of $K(X)$ of index $d$. 
For genus $1$ curves, the gonality is the maximum of $2$ and the index of the curve. If the gonality $\gamma$ is at least $3$, the result above plus Riemann-Roch show there are infinitely many morphisms of degree $\gamma$ from $X$ to the projective line. 
