In the context of the quoted proposition arises the following question.
Let $X$ be an abstract nonsingular curve (as defined in p. 42), $P \in X$ and let $\phi: X-P \rightarrow \mathbb{P}^n$ a morphism. The proof of the proposition argues that we can assume that $\phi(X-P) \cap U \neq \emptyset$, where $U$ is the open set of $\mathbb{P}^n$, where all homogeneous coordinates $x_0,x_1,\dots,x_n$ are non-zero. The argument is inductive and boils down to showing that this is true for $n=1$ and the details of this are left to the reader. So let's assume that $\phi : X-P \rightarrow \mathbb{P}^1$ and suppose that $\phi(X-P) \cap U = \emptyset$, $U$ being the open set of $\mathbb{P}^1$ where both $x_0,x_1$ are non-zero. Then $\phi(X-P) \subset \left\{(1,0),(0,1)\right\}$. Since $\phi$ is continuous and $\phi(X-P)$ irreducible we must have without loss of generality that $\phi(X-P) = \left\{(1,0)\right\}$. But now i don't see how to arrive at a contradiction. Any ideas?
The base case of the induction is $n=0$. Your example is to have $\phi$ map all of $X-P$ to a single point, $(1,0)\in\mathbb{P}^1$. But we can also think of this single point as a copy of $\mathbb{P}^0$, so we have reduced to looking at a map $X-P\to\mathbb{P}^0$.
But for $n=0$, the proposition, which says that we can uniquely extend $\phi$ to all of $X$, is obvious—we have no choice but to map all of $X$ to the single point of $\mathbb{P}^0$.