We consider the so called 'Steiner construction', which can be looked up e.g. in this script e.g. which gives us an irreducible curve.
The construction works as follows: Let say we have $\Lambda_1, ..., \Lambda_n \cong \mathbb{P}^{n-2} \subset
\mathbb{P}^{n} $ codimension two subspaces in
$\mathbb{P}^{n} $. Let for every $i =1,...,n$ be $\{H_t^i \}_{t \in \mathbb{P}^{1}}$
the pencil of hyperplanes containing the plane $\Lambda_i$
with
additional assumption that for every $t_0 \in \mathbb{P}^{1}$ the planes
$H_{t_0}^1,...H_{t_0}^n$ are independent, i.e., intersect in a unique point $p(t_0)$.
Consider the union of intersections of these hyperplanes:
$$ C:= \bigcup_{t \in \mathbb{P}^{1}} H_t^1 \cap ... \cap H_t^n $$
In the linked script is is claimed that this curve is a rational normal curve, i.e. projectively equivalent to the curve obtained as image of the Veronese map
$$ v_d: \mathbb{P}^1 \to \mathbb{P}^n, [X_0:X_1] \mapsto [X_0^n: X_0^{n-1}X_1:..., X_1^n]$$
In this question I wanted to discuss how to prove that the Steiner construction above gives indeed a rational normal curve defined in more 'conventional' sense as Veronese's image.
Here I want to prove a weaker statement inspired by a user's comment
which is unfortunately meanwhile removed.
Question/Problem: How to show that this curve
$C= \bigcup_{t \in \mathbb{P}^{1}} H_t^1 \cap ... \cap H_t^n \subset \mathbb{P}^n$
constructed above is non-degenerated, that means there is no
hyperplane $\mathbb{P}^{n-1} \cong H \subset \mathbb{P}^n$, which contains $C$? If we clearly already know that $C$ ration normal curve, then it's clear, but let's play only with Steiner's construction as definition of $C$.
The problem is that the spaces $\Lambda_1, ..., \Lambda_n \cong \mathbb{P}^{n-2} \subset \mathbb{P}^{n}$ in the construction seem to be choosen arbitrary and that makes it hard for me to prove the claim.
A simple case I can solve myself is if the $\Lambda_1, ..., \Lambda_n \subset \mathbb{P}^{n}$ can be choosen as follows: Let $p_1,..., p_n \in C$ be points in general position, that is the linear closure $\overline{p_1,..., p_n}$ is a hyperplane $H$, that is a linear subspace of codimension one in $\mathbb{P}^n$.
Take $\Lambda_i := \overline{p_1,..., \hat{p_i},..., p_n}$ where $\hat{p_i}$ mean that this point was removed. Then one concludes that if there exist a hyperplane $H$ which contains $C$ then it is equals to $\overline{p_1,..., p_n}$. But then by construction it is easy to see that if we chose $t \in \mathbb{P}^1$ such that for all $i=1,..., n$ the hyperplane $H^i_t \neq H$, then the intersetion of $H_t^1 \cap ... \cap H_t^n$ with $H$ is empty, therefore $H_t^1 \cap ... \cap H_t^n \not \in H$.
But the Steiner construction assumes
$\Lambda_1, ..., \Lambda_n \cong \mathbb{P}^{n-2} \subset
\mathbb{P}^{n}$ to be pairwise distinct but otherwise arbitrary.
How can I show that non-degeneracy of
$C= \bigcup_{t \in \mathbb{P}^{1}} H_t^1 \cap ... \cap H_t^n \subset \mathbb{P}^n$
in that case?
My idea was to find enough points $p_1,..., p_n$ with
$p_i \in \Lambda_i$ which are in general position and then as above
to show that there is a point $c \in C$ which isn't contained
in $\overline{p_1,..., p_n}$.
Problem: Can I such points always be found? Are results/estimations known how many points of $C$ should a $\Lambda_i$ contain? How can I show that every $\Lambda_i$ contains a point $p_i$ such that $p_1,..., p_n \in C$ are in general position, i.e. span a hyperplane in $\mathbb{P}^{n}$?