Base-point-freeness.
A rational map $C\dashrightarrow V\subset\mathbb P^n$ from a curve $C$ to a projective variety $V$ is defined at all the smooth points. In particular, taking $V=\mathbb P^n$, any linear series $f:C\dashrightarrow\mathbb P^n$ on a smooth projective curve is in fact base-point-free. Recall that the locus where $f$ is not defined coincides with the base locus of the corresponding linear series; but the latter is defined as the intersection of all the divisors in the system, namely $$\textrm{Bl}(f)=\bigcap_{H\subset\mathbb P^{n}}f^{-1}(H),\,\,\,\,\,\,\,\,\,\,\,\,\,\,H\,\,\textrm{all the hyperplanes in}\,\,\mathbb P^n,$$ and when $n=1$, a point $P\in C$ cannot satisfy $f(P)=x$ for all $x\in \mathbb P^1$, so in the case you are interested in the base locus is empty for this very concrete reason.
Completeness.
Let $\textrm{Pic}^{e}(C)$ denote the group of degree $e$ line bundles on $C$.
As you said, the gonality being $d$ means that any $L\in \textrm{Pic}^{d-1}(C)$ is such that any vector subspace $U\subseteq H^0(C,L)$ has $\dim U\leq 1$. But we have a $g^1_d$, namely a couple $(\mathscr L,V)$ where $\mathscr L\in\textrm{Pic}^{d}(C)$ and $V\subseteq H^0(C,\mathscr L)$ has dimension 2. We want to show that in fact $h^0(C,\mathscr L)=2$.
Let us take any point $P\in C$. Then, $L=\mathscr L(-P)\in \textrm{Pic}^{d-1}(C)$, so by what we said we must have $h^0(C,L)\leq 1$. But the inclusion $H^0(C,L)\subset H^0(C,\mathscr L)$ produces, in general, either no dimension jump, or a jump of one dimension. This means that $h^0(C,\mathscr L)\leq 2$, hence it is in fact $=2$.
Clearly any base point of $|2P_0|$ would have to be $P_0$. Since the linear system $|2P_0|$ is $1$-dimensional, there exists another effective divisor $D \neq 2P_0$, linearly equivalent to $2 P_0$. But as $P_0$ is a base point we have $D = P_0 + Q$ for some other point $Q \in X$. And hence $Q \sim P_0$, which is a criterion for rationality (example II 6.10.1).
Best Answer
Let $C$ be a smooth projective curve of genus $g$ over an algebraically closed field.
Let $p_1,\cdots, p_n$ be $n$ points on $C$.
Then if you consider the divisor $D=p_1+\cdots+ p_n$, the linear system $\mathcal E\subset |D|$ consisting of the divisors $E\in |D|$ with $p_1$ in their support is tautologically a (non-complete) linear system having $p_1$ as a base point.
If $n\geq 2g$ that linear system $\mathcal E$ is a hyperplane in the projective space $|D|$ of dimension $n-g$ and has thus dimension $n-g-1$ .
Note that $|D|$ itself has no base point if $n\geq2g$.
All this follows from Riemann-Roch but if you haven't learned this theorem yet you can accept the above on faith for the moment if you trussst in meee.