Degree of map to $\mathbb{P}^1$ bounded by degree of divisor

Question

Let $X$ be a curve of genus $g$. Suppose $D$ is a divisor with $\dim H^0(D)\geq 2$. Then there exists a non-constant $f\in H^0(D)$. In the top answer to this question, Hartshorne 4.1.6 Gonality of a curve, it is stated that $f$ gives a finite morphism $f:X\to\mathbb{P}^1$ of degree at most $\deg D$. My question is: Why is the degree of $f$ at most $\deg D$?

Answer 1

If $|D|$ is base point free, $\deg(f) = \deg(D)$. Otherwise, if $P$ is a base point of $|D|$, the map given by $|D-P|$ is the same as the map $f$, therefore by induction $\deg(f) \le \deg(D-P) = \deg(D) - 1$.