Understanding Norms on Vector Spaces Let $\|\cdot\|$ be a norm (not necessarily the standard norm) on $\mathbf R^2$ and $S$ be the set of all the vectors $v$ such that $\|v\|=1$.
For any point $p\in S$, let $\ell_p$ denote the line joining origin and $p$.
I think the following should hold:

For any point $p\in S$, there is a neighborhood $U$ of $p$ such that no line parallel to $\ell_p$ intersects $U\cap S$ in more than one point.

This looks visually obvious, but I am facing difficulty proving this.
Can anyone please help?
Thanks.
 A: Lemma 1: Let $A$ and $B$ be two points having norm $1$. Then theere is no point on the line segment joining $A$ and $B$ which has norm greater than $1$.
Proof: The norm is a convex function because $\|tx+(1-t)y\|\leq t\|x\|+(1-t)\|y\|$ for all points $x$ and $y$ and $t\in [0, 1]$.
Lemma 2: If there exist three points $A, B$ and $C$ having norm $1$, then all points inside the triangle $ABC$ have norm strictly less than $1$.
Proof: Suppose not. Then there is a point $D$ inside $ABC$ such that $D$ has norm $1$. Join $D$ with the origin and extend until we cut a side of $ABC$ at a point say $X$. Then $X$ has norm greater than $1$ but this is not possible by Lemma 1. 
Now we come to the problem. Assume the contrary to the proposition and let $p$ be a point having norm $1$ such that every neighborhood of $p$ has two distinct points lying on a line parallel to $\ell_p$ such that both these points have norm $1$.
Consider this diagram:https://www.dropbox.com/s/om84ixeib7l943r/IMG_20150602_184748.jpg?dl=0
Let $m$ be a line parallel to $\ell_p$ having two points $q$ and $r$ such that norm of both $q$ and $r$ is $1$.
Note that by Lemma 2 we know that all points inside the triangle $pqr$ have norm strictly less than $1$.
Suppose we have a line $\ell$ (unmarked in the figure) between $\ell_p$ and $m$ and parallel to $\ell_p$ on which lie two points $s$ and $t$ both having norm $1$.
There are two possibilities.
In possibility 1 as shown in the figure, first note that we could choose $\ell$ such that $s$ and $t$ lie in a neighborhood of $p$ which is small enough such that the line joining the origin with $s$ has a point $s'$ which lies inside $pqr$. Now norm of $s'$ is greater than $1$, but this is not possible by Lemma 2.
In possibility 2, we have a triangle $ptr$ whose vertices all have norm $1$ inside of which lies a point $s$ which also has norm $1$. This again is not possible by Lemma 2.
