When the intersection of projective varieties is finite What I want to solve is

Let $C$ be a projective curve, $P\in C$. Then there is a binational morphism $f:C \longrightarrow C'$, $C'$ a projective plane curve, such that $f^{-1}(f(P))=\{P\}$.

Here, a projective curve means that $C\subseteq \mathbb{P}^{n+1}$ is projective variety whose dimension is 1, and projective "plane" curve is just a curve in $\mathbb{P}^2$. The outline of proof is following :
We can assume : Let $T,X_1,\cdots,X_n,Z$ be coordinates for $\mathbb{P}^{n+1}$; Then $C\cap V(T)$ is finite ; $C\cap V(T,Z)=\emptyset$; $P=[0:0:\cdots :0:1]$ and $k(C)$ (function field on C) is algebraic over $k(u)$, where $u=\bar{T}/\bar{Z}\in k(C)$.
But I can't understand this assumption, because basically I think I'm not sure when the intersection of two varieties is finite set.(In $\mathbb{P}^2$, I know this set is finite if the homogeneous polynomials have no common component, but I don't know the case for larger dimension) Can anyone explain why we can assume such things?
 A: $C\cap V(T)$ is a closed subset of $C$, which means it's either all of $C$ or a finite collection of points. This is developed in Hartshorne chapter I exercise 4.8, for instance. So if $C$ isn't contained in $V(T)$, then this intersection is just a finite collection of points; if $C\subset V(T)$, just go to the $\Bbb P^{n-1}$ which is $V(T)$, pick new coordinates, and try again.
Your larger question about when the intersection of two projective varieties is finite is more difficult. The best available thing here is that if $X,Y$ are closed subvarieties of $\Bbb P^n$ of dimension $r,s$ respectively, then every irreducible component of $X\cap Y$ has dimension at least $r+s-n$, and if $r+s-n\geq 0$ then this intersection is nonempty. This is only an existence result - you can certainly have a nonfinite intersection if $r+s-n <0$: think about two lines in $\Bbb P^3$, for instance. They can be the same line, intersect in a point, or be skew lines and have no intersection. But the intersection can only be finite if $r+s\leq n$.
