= Solution
A <divisor on a complex manifold>, specialized to a compact <Riemann surface>, is a finite sum $D=\sum_p n_pp$ with integer coefficients, and $\deg D=\sum_pn_p$. Two divisors are linearly equivalent when $D'=D+(h)$ for a nonzero <meromorphic function> $h$, whose <principal divisor on a complex manifold> records zero orders minus pole orders. Define
$$
L(D)=\{0\}\cup\{f\text{ meromorphic}:(f)+D\ge0\},\qquad \ell(D)=\dim_{\mathbb C}L(D).
$$
Multiplication by $1/h$ gives an isomorphism $L(D)\to L(D')$, because $(f/h)+D'=(f)+D$. Its inverse is multiplication by $h$, proving $\ell(D')=\ell(D)$.
On the <Riemann sphere> move $P$ to infinity by a Möbius coordinate. Meromorphic functions on the sphere are rational; those with no finite poles and a pole of order at most one at infinity are exactly $az+b$. Thus \b[$\ell(P)=2$], without using Riemann–Roch.
The <Riemann-Roch theorem> for a compact connected genus-$g$ surface is $\ell(D)-\ell(K-D)=\deg D+1-g$, where $K$ is a canonical divisor. Choose $D=(g+1)P$. Since the second dimension is nonnegative, $\ell(D)\ge2$. Constants span only one dimension, so a nonconstant $f$ exists whose only possible pole is at $P$ with order at most $g+1$. The degree of its holomorphic map to the sphere is the total pole order. Therefore
$$
\boxed{\deg(f:C\to S^2)\le g+1}.
$$
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