[proofplan]
The complete linear system $|D|$ — the set of effective divisors linearly equivalent to $D$ — is parametrised by the projective space $\mathbb{P}(\mathcal{L}(D))$ via $[f] \mapsto \operatorname{div}(f) + D$. When $\ell(D) = 1$, the projective space $\mathbb{P}(\mathcal{L}(D))$ is a single point, so $|D|$ has exactly one element. Concretely: existence comes from the divisor-of-zeros map applied to any nonzero $f \in \mathcal{L}(D)$; uniqueness comes from the fact that any two nonzero elements of a one-dimensional space are scalar multiples, and scalar multiplication does not change a principal divisor.
[/proofplan]
[step:Pick a nonzero generator of $\mathcal{L}(D)$ and produce an effective representative]
Since $\ell(D) = \dim_k \mathcal{L}(D) = 1$, the $k$-vector space $\mathcal{L}(D)$ is one-dimensional. Choose a nonzero element $f \in \mathcal{L}(D)$; every nonzero element of $\mathcal{L}(D)$ is then of the form $\lambda f$ for some $\lambda \in k^\times$.
Define
\begin{align*}
D_f := \operatorname{div}(f) + D \in \mathrm{Div}(C).
\end{align*}
By the defining condition of $\mathcal{L}(D)$, the divisor $\operatorname{div}(f) + D$ is effective ($\geq 0$); hence $D_f$ is effective. Moreover $D_f - D = \operatorname{div}(f)$ is a principal divisor, so $D_f \sim D$.
This produces an effective divisor linearly equivalent to $D$, establishing existence.
[/step]
[step:Show that any effective representative arises from a nonzero element of $\mathcal{L}(D)$]
Let $D'$ be any effective divisor with $D' \sim D$. By definition of linear equivalence, there exists $g \in k(C)^\times$ with
\begin{align*}
D' - D = \operatorname{div}(g),
\end{align*}
i.e., $D' = \operatorname{div}(g) + D$. Effectivity of $D'$ rewrites as
\begin{align*}
\operatorname{div}(g) + D \geq 0,
\end{align*}
which is exactly the defining condition for $g \in \mathcal{L}(D)$. Since $g \in k(C)^\times$ is nonzero, $g \in \mathcal{L}(D) \setminus \{0\}$.
Thus every effective $D' \sim D$ is of the form $\operatorname{div}(g) + D$ for some nonzero $g \in \mathcal{L}(D)$.
[/step]
[step:Use one-dimensionality of $\mathcal{L}(D)$ to conclude uniqueness]
By Step 2, any effective $D' \sim D$ is $D' = \operatorname{div}(g) + D$ for some nonzero $g \in \mathcal{L}(D)$. By Step 1, $\mathcal{L}(D) = k \cdot f$, so $g = \lambda f$ for some $\lambda \in k^\times$. The principal divisor of a nonzero scalar multiple equals the principal divisor of the function:
\begin{align*}
\operatorname{div}(\lambda f) = \operatorname{div}(\lambda) + \operatorname{div}(f) = 0 + \operatorname{div}(f) = \operatorname{div}(f),
\end{align*}
since $\lambda \in k^\times$ is a nonzero constant on $C$ and constants have zero principal divisor on a smooth projective curve. Therefore
\begin{align*}
D' = \operatorname{div}(g) + D = \operatorname{div}(\lambda f) + D = \operatorname{div}(f) + D = D_f.
\end{align*}
This holds for every effective representative, so $D_f$ is the unique effective divisor linearly equivalent to $D$.
[guided]
We have shown two things:
\begin{enumerate}
\item Existence (Step 1): the divisor $D_f := \operatorname{div}(f) + D$ is effective and linearly equivalent to $D$.
\item Reduction (Step 2): every effective $D' \sim D$ has the form $D' = \operatorname{div}(g) + D$ for a nonzero $g \in \mathcal{L}(D)$.
\end{enumerate}
The remaining task is to argue that all such $g$ produce the *same* effective divisor as $f$ does. This is where the hypothesis $\ell(D) = 1$ enters: it says $\mathcal{L}(D)$ is a one-dimensional $k$-vector space, so any two nonzero elements are scalar multiples of each other.
Concretely, $g \in \mathcal{L}(D) \setminus \{0\}$ and $f \in \mathcal{L}(D) \setminus \{0\}$ both lie in the one-dimensional space $k \cdot f$; the unique linear combination expressing $g$ in terms of $f$ has the form $g = \lambda f$ with $\lambda \in k$, and $\lambda \neq 0$ because $g \neq 0$.
Why does scaling not change the principal divisor? The principal divisor is defined by $\operatorname{div}(h) = \sum_p \operatorname{ord}_p(h) \cdot p$, where $\operatorname{ord}_p$ is the discrete valuation at $p$. A nonzero constant $\lambda \in k^\times$ is a unit in every local ring $\mathcal{O}_{C,p}$, hence $\operatorname{ord}_p(\lambda) = 0$ for all $p$, so $\operatorname{div}(\lambda) = 0$. Multiplicativity of $\operatorname{ord}_p$ gives $\operatorname{ord}_p(\lambda f) = \operatorname{ord}_p(\lambda) + \operatorname{ord}_p(f) = \operatorname{ord}_p(f)$, hence $\operatorname{div}(\lambda f) = \operatorname{div}(f)$.
Combining these:
\begin{align*}
D' = \operatorname{div}(g) + D = \operatorname{div}(\lambda f) + D = \operatorname{div}(f) + D = D_f.
\end{align*}
So *any* effective $D' \sim D$ equals $D_f$. The effective representative is unique.
\textbf{Why one-dimensionality is essential.} If $\ell(D) \geq 2$, the space $\mathcal{L}(D)$ contains two linearly independent elements $f_1, f_2$; the divisors $\operatorname{div}(f_1) + D$ and $\operatorname{div}(f_2) + D$ are both effective and equivalent to $D$, but they differ — they are distinct points of the projective space $\mathbb{P}(\mathcal{L}(D)) \cong \mathbb{P}^{\ell(D) - 1}$. The complete linear system $|D|$ is parametrised by this projective space, and it reduces to a single point exactly when $\ell(D) = 1$.
[/guided]
[/step]
