I'm trying to understand a step in Harris book in proof of a proposition (Algebraic Geometry by Joe Harris, p. 230) which looks quite confusing:
Proposition 18.10. Let $X \subset \mathbb{P}^n$ be an irreducible nondegenerate variety of dimension $k \ge 1$, and let $Y = X \cap H \subset H \cong \mathbb{P}^{n-1}$ be a general hyperplane section of $X$. Then $Y$ is nondegenerate in $\mathbb{P}^{n-1}$, and if $k \ge 2$ then $Y$ is irreducible as well
PROOF. To begin, the nondegeneracy statement is relatively elementary; we will prove it for any $H$ intersecting $X$ generically transversely. Suppose that $Y$ is contained in a subspace $\Lambda \cong \mathbb{P}^{n-2}$, and let $\{H_{\lambda} \}$ be the family of hyperplanes in $\mathbb{P}^{n}$ containing $\Lambda$. Since $X$ is not contained in any finite union of hyperplanes, it will intersect a general member $H_{\lambda}$, of this pencil in at least one point $P$ not lying on $Y$; the point $P$ will lie on an irreducible component $Z_{\lambda}$ $X \cap Z_{\lambda}$, not contained in $Y$. Then...
Problem: I not understand the implication that 'since $X$ is not contained in any finite union of hyperplanes, it will intersect a general member $H_{\lambda}$, of this pencil in at least one point $P$ not lying on $Y$'.
I think that this implication might be wrong, because if we call
statement A :='$X$ is not contained in any finite union of hyperplanes ' and
statement B := 'variety $X$ intersect a general member $H_{\lambda}$, of this pencil in at least one point $P$ not lying on $Y$'
then we run into troubles. Recall 'general member' means that there exist an open $U \subset \mathbb{P}^1$ such that for every $\lambda \in U$ the intersection $X \cap H_{\lambda_U}$ is not contained in $Y$.
The point is why A implies B? More concretly why statement A forbids the existence
of an infinite subset $S \subset \mathbb{P}^1$ with the same cardinality like
naturals $\vert S \vert = \vert \mathbb{N} \vert$ such that for every memeber
$\lambda_s \in S$ the intersection $X \cap H_{\lambda_s}$ is contained in $Y$.
Then such $S$ would be dense in $\mathbb{P}^1$ and B wrong since the 'general member'
property means that there exist open $U \subset \mathbb{P}^1$ such that for every
$\lambda \in U$ the intersection $X \cap H_{\lambda_U}$ is not contained in $Y$.
Question: why statement A implies that such subset $S$ cannot exist?