In the newly updated Rising Sea (August 2022), this exercise is now 19.4.M. And in this version, there is a new exercise preceding it,
19.4.K. EXERCISE. If $\mathcal{F}$ is a torsion sheaf on $C$, show that $\deg(\mathcal{F})\geq 0$, with equality if and only $\mathcal{F}=0$.
Here $C$ is an integral projective curve. From this, one can deduce that if a line bundle $\mathcal{N}$ has a nonzero global section, and $\mathcal{N}$ is not trivial, then $\deg(\mathcal{N})>0$. This is done by examining the exact sequence
$$0\to \mathcal{O}_C\to \mathcal{N}\to \mathcal{F}\to 0$$
where the first map is the one induced by the global section of $\mathcal{N}$, and $\mathcal{F}$ is defined to be the cokernel. In this section, it is shown that both rank and degree are additive on exact sequences. The first gives that the rank of $\mathcal{F}$ is $0$, and the second gives that the degrees of $\mathcal{F}$ and $\mathcal{N}$ are equal. By 19.4.K, using the fact that $\mathcal{N}$ is not trivial, we have $\deg(\mathcal{N})>0$.
This then shows that for $\mathcal{N}$ very ample, we have $\deg(\mathcal{N})>0$. But we want this for ample line bundles, not very ample line bundles. As is mentioned in the comments to Andrew's answer, $\deg(\mathcal{L}^{\otimes n})=n\deg(\mathcal{L})$ (even better, degree is additive across tensor products), even for singular curves. But it seems like this may take a lot of effort, so we will go around this slightly.
Suppose $\mathcal{L}$ is ample, with $\mathcal{L}^{\otimes n}$ very ample (or even just has basepoint-free) for $n\geq N$, and $\mathcal{M}$ is any line bundle. Set $\mathcal{N}:=\mathcal{L}^{\otimes n}$ for some $n\geq N$. Tensoring the above exact sequence with $\mathcal{M}$ gives
$$0\to \mathcal{M} \to \mathcal{N}\otimes \mathcal{M}\to \mathcal{F}\otimes \mathcal{M}\cong \mathcal{F}\to 0$$
exact, from which we get
$$\deg(\mathcal{N}\otimes \mathcal{M})=\deg(\mathcal{M})+\deg(\mathcal{F})=\deg(\mathcal{M})+\deg(\mathcal{N})$$
Plugging in $\mathcal{N}=\mathcal{L}^{\otimes n}$, and setting $\mathcal{M}=\mathcal{L}^{\otimes a}$, we see that
$$\deg(\mathcal{L}^{\otimes n}\otimes \mathcal{L}^{\otimes a})=\deg(\mathcal{L}^{\otimes n})+\deg(\mathcal{L}^{\otimes a})$$
for $n>N$, and any $a\in\mathbb{Z}$. And then, for any $a,b\in\mathbb{Z}$, choosing $n$ so that $n,n+a\geq N$, we have
$$\deg(\mathcal{L}^{\otimes n})+\deg(\mathcal{L}^{\otimes a})+\deg(\mathcal{L}^{\otimes b})=\deg(\mathcal{L}^{\otimes (n+a+b)})=\deg(\mathcal{L}^{\otimes n})+\deg(\mathcal{L}^{\otimes (a+b)})$$
which implies $\deg(\mathcal{L}^{\otimes a})+\deg(\mathcal{L}^{\otimes b})=\deg(\mathcal{L}^{\otimes (a+b)})$, for any $a,b\in\mathbb{Z}$, and so $0<\deg(\mathcal{L}^{\otimes N})=N\deg(\mathcal{L})\implies \deg(\mathcal{L})>0$.
