Consider a rank two vector bundle $\mathcal{E}$ over $\mathbb{P}^2$, set $X = \mathbb{P}(\mathcal{E})$, and let $\pi:X\rightarrow\mathbb{P}^2$ be the projection. I would like to show that $X$ contains a surface $W\subset X$ such that $\pi_{|W}:W\rightarrow\mathbb{P}^2$ is generically one to one, and $W$ contains at most finitely many fibers of $\pi:X\rightarrow\mathbb{P}^2$.
Let $L\subset\mathbb{P}^2$ be a line and $\mathcal{U} := \mathbb{P}^2\setminus\{L\}\cong\mathbb{A}^2$. Then there is an isomorphism $\phi:\mathcal{U}\times\mathbb{P}^1\rightarrow X|_{\mathcal{U}}$. Take two general polynomials $f,g:\mathcal{U}\rightarrow K$, where $K$ is the base field, and consider the function $s:\mathcal{U}\rightarrow \mathcal{U}\times\mathbb{P}^1$ given by $s(x) = (x,f(x),g(x))$. Then $\phi\circ s:\mathcal{U}\rightarrow X|_{\mathcal{U}}$ is a section of $X|_{\mathcal{U}}\rightarrow\mathcal{U}$.
Set $S = (\phi\circ s)(\mathcal{U})\subset X|_{\mathcal{U}}\subset X$. Then $S$ is irreducible. Let $W$ be the closure of $S$ in $X=\mathbb{P}(\mathcal{E})$. Then $W$ is irreducible as well. Now, $W$ might contain fibers of $\pi$ over $L$. However, if $W$ contains infinitely many fibers of $\pi$ over $L$, then it contains $\pi^{-1}(L)$ and hence $\pi^{-1}(L)$ would be a component of $W$ contradicting the irreducibility of $W$.
So $W$ contains at most finitely many fibers of $\pi$. Is this argument correct?
Thank you.
Question: "Let $L⊂\mathbb{P}^2$ be a line and $U:=\mathbb{P}^2∖{L}≅\mathbb{A}^2$. Then there is an isomorphism $ϕ:U×\mathbb{A}^2→X|U$."
Answer: If $E$ is locally trivial on $S:=\mathbb{P}^2$ it follows for any open subscheme $U \subseteq S$ that $\pi^{-1}(U):=U\times_S \mathbb{P}(E^*)\cong \mathbb{P}((E_U)^*)$ and since the restriction $E_U\cong U\{e_0,e_1\}$ is trivial of rank 2, it follows $\pi^{-1}(U) \cong Proj(Sym^*_U(U\{x_0,x_1\})):=Proj(A[x_0,x_1]) \cong U\times_k \mathbb{P}^1$ where $U:=Spec(A), A:=k[x,y]$. Hence your claim that $\pi^{-1}(U) \cong U \times_k \mathbb{A}^2$ cannot be true.