Understanding Norms On a Vector Space (Part II) This question is motivated by a previous question of mine.
Let $\|\cdot \|$ be any norm(not necessarily the standard norm) on $\mathbf R^3$ and $S$ be the set of all the points with norm $1$.
Let $p$ be point in $S$ and $\ell_p$ be the line joining the origin and $p$.

Then there exists a neighborhood $U$ of $p$ in $\mathbf R^3$ such that no line parallel to $\ell_p$ intersects $U\cap S$ in more than $1$ point.

The link given above discusses the same problem in $\mathbf R^2$, where I have provided an answer too. The reasoning given there does not extend to this case.
In fact, there is no reason to think that the proposition is false for any $n$.
My ultimate motive behind this is to show that $S$ is a topological manifold.
 A: Let $n$ denote the norm in question, since we are in finite dimensions, it is
equivalent to the usual Euclidean norm. (Presumably by parallel, you meant parallel in the Euclidean sense.)
Let $p \in S \subset \mathbb{R}^n$, that is, $n(p)=1$.
Let $\pi$ be the orthogonal projection onto the subspace $\{x | \langle p, x \rangle = 0 \}$. Since $n \circ \pi$ is continuous, the
set $U = (n \circ \pi)^{-1}((-{1 \over 2} , {1 \over 2})) \cap \{x | \langle p, x \rangle > 0 \}$ is open and contains $p$.
If $L$ is a line parallel to $l_p$, then we can write $L$ as the
range of
$\lambda(t) = \delta + t p$, where $\delta \bot p$.
Suppose $t$ is such that $\lambda(t) \in U$, then
$n(\pi(\lambda(t))) = n(\delta) < {1 \over 2}$, and
$\langle p, \lambda(t) \rangle > 0$, hence $t >0$.
Now suppose $L$ intersects $S \cap U$ in two places, then there are
$0<t_1<t_2$ such that $n(\lambda(t_1)) = n(\lambda(t_2)) = 1$ and
$\lambda(t_k) \in U$.
The function $\phi(t)=n(\lambda(t))$ is convex, $\phi(t_1) = \phi(t_2) = 1$, hence we have $\phi(t) \ge 1$ for all $t \notin (t_1,t_2)$. In particular,
we have $\phi(0) = n(\lambda(0)) = n(\delta) \ge 1$, a contradiction. Hence
$L$ can intersect $U \cap S$ in at most one point.
