Problem I.3.3(c) from Hartshorne says that if $\phi: X \rightarrow Y$ is a morphism of varieties and if $\phi(X)$ is dense in $Y$ then the homomorphism of local rings $\phi_P^* : \mathcal{O}_{Y,\phi(P)} \rightarrow \mathcal{O}_{X,P}$ is injective for every $P \in X$.
Here is how i tried to prove that:
Fact 1: Take $P \in X$ and let $\langle V, f \rangle$ be a non-zero germ at $\phi(P)$. That means that for every neighborhood $W$ of $\phi(P)$, $f$ is non-zero on $W \cap V$. Suppose now that $\phi_P^*(\langle V, f \rangle) = 0$. By definition of $\phi_P^*$ (if i understand correctly) this means that $\langle \phi^{-1}(V), f \circ \phi \rangle = 0$, and this means that there exists a neighborhood $U$ of $P$ such that $f \circ \phi$ is zero on $\phi^{-1}(V) \cap U$.
Fact 2 (edited after comments): Since $V$ is a neighborhood of $\phi(P)$ and $f$ is non-zero on $V$, there exists some $\xi \in V$ such that $f(\xi) \neq 0$. Since $\phi(X)$ is dense in $Y$, we must have that $\xi$ is a limit point of $\phi(X)$, i.e. for every neighborhood $W$ of $\xi$ we have $\phi(X) \cap W \neq \emptyset$.
Question 1: Is there a way to combine Facts 1 and 2 to arrive at a proof?
Question 2: I found some hints online saying very concisely that "if $\phi_P^*(f)=0$, then $f$ vanishes on the dense set $V \cap \phi(X)$ and so $f$ must be zero on $V$." I do see why $V \cap \phi(X)$ is dense in $V$ but i don't see why $f$ vanishes on $V \cap \phi(X)$. (In the hint there is no mention of what $V$ is, i assume it is some neighborhood of $\phi(P)$.) I would like an explanation of this hint.
PS: An instructive answer to question 2 will be enough.