Doubts about proof of Castelnuovo theorem IV, 6.4 of Hartshorne

In the proof of Theorem 6.4 (Castelnuovo) of Hartshorne, chapter IV, it reads: "to show that $P_i$ is not a base point of the linear system $|nD-P_1-\dots-P_{i-1}|$ it is sufficient to find a surface of degree $n$ in $\mathbb{P}^3$ containing $P_1,\dots,P_{i-1}$ but not $P_i$".

Why? Maybe I didn't get the geometric interpretation of what a base point is?

If $P_i$ is a base point of a linear system $\mathfrak{d}$, that means that every element of $\mathfrak{d}$ vanishes on $P_i$. So finding a surface in $\Bbb P^3$ which doesn't pass through $P_i$ means the equation of that surface doesn't vanish on $P_i$, and that gives you an element of $\mathfrak{d}$ which doesn't vanish on $P_i$.

Consider the following exact sequences of sheaves: $$0\to\mathcal{I}_X(n)\to\mathcal{O}_{\Bbb P^3}(n)\to \mathcal{O}_X(n)\cong\mathcal{O}_X(nD)\to 0$$ $$0\to \mathcal{O}_X(nD-P_1-\cdots-P_{i-1})\to \mathcal{O}_X(nD)\to \bigoplus_{j=1}^{j=i-1} \mathcal{O}_{P_i}\to 0.$$ A surface of degree $n$ is a global section of $\mathcal{O}_{\Bbb P^n}(n)$, which as long as it doesn't contain $X$ gives a nontrivial global section of $\mathcal{O}_X(nD)$. If this global section vanishes at all the $P_j$, then it's a global section of $\mathcal{O}_X(nD-P_1-\cdots-P_{i-1})$ by taking global sections of the second sequence as global sections is left exact.