A curve of genus $g\geq 2$ has a closed point of degree at most $2g-2$ over base field. I am working on the following problem

[R. Vakil] Exercise 19.8.B: Suppose $C$ is a curve of genus $g>1$ over
  a field $k$ that is not algebraically closed. Show that $C$ has a
  closed point of degree at most $2g-2$ over the base field.

I have no idea how to do this question. This is what I know: Since $g>1$, so the sections of the dualising sheaf $\omega$ defines a morphism $\varphi: C\rightarrow\mathbb{P}^{g-1}$. Hence it defines a morphism $\varphi:C\rightarrow C'$ from $C$ onto its image curve $C'$. The degree of this morphism is $2g-2$, which corresponds to the degree of the extension of the function field $[k(C):k(C')]$. 
How should I go on from here? i.e. Where to find that closed point? I tried to break down the cases where either $\varphi$ is a closed embedding or it is hyperelliptic but they don't seem to help.
 A: Okay, by hypothesis $\omega$ is degree $2g -2$ and it has a global section $s \in H^0(\omega)$.  
Then $\omega$ is $\mathcal O(D)$ for the divisor $D = \sum_p \nu_p(s) \cdot p$, where the sum is taken over all points $p$.  And $2g -2 =\deg \omega = \sum_p \nu_p(s) \cdot \deg p$, and all the $v_p(s)$ are positive since $s$ is global.  
So worst-case scenario, we have point of degree $2g -2$.  
A: Pick a point $p\in C$. Consider invertible sheaf $\omega_C (-p)$. In the process of exercise 19.8.A one shows using Riemann-Roch that $h^0 (C, \omega(-p))=g-1$ and so under assumption $g>1$ it has a global section $s$. By the description of exercise 14.2.J, $s$ corresponds to a rational section $t$ of  $\omega_C $, s.t. $div (t) - p \geq 0$. By Hartog's lemma, a section without poles lifts to a global section $t$ of $\omega_C $ with a zero at $p$. Now the degree of $\omega_C $ is $2g-2$ and the formula $2g -2 = \sum_q val_q(t) \cdot \deg q$ with $val_q(t) \geq 0$ shows $deg \ p \leq 2g-2$, since for $p$,  $val_p(t) > 0$.
