Assume that $(X,\|\cdot\|)$ is a Banach space with $\|\cdot\|$ strictly convex. Define $S=\{x\in X:\|x\|=1\}$. Suppose that $\varepsilon>0$ and $x_0\in S$ and define $$ B_\varepsilon=\{x\in X:\|x-x_0\|\le\varepsilon\}\cap S. $$ Consider the following problem:
Can we find $y\in X$ such that $\|y\|<1$ and any line which contains $y$ must intersect $B_\varepsilon$?
Now consider the following possible solution:
Take a supporting hyperplane $H$ for the convex set $B=\{x\in X:\|x\|\le 1\}$ at the point $x_0$. Once $\|\cdot\|$ is strictly convex, one can find $z\in X\setminus\{0\}$ such the set $$ H_z=\{h+z:h\in H\}, $$ has the property that $H_z\cap B_\varepsilon\neq \emptyset$ and $H_z\cap B_\varepsilon=H_z\cap S$. Now let $Q$ be the region enclosed by $H_z$ and $B_\varepsilon$. It follows that if $y\in Q$, then any line which contains $x$ must intersect $B_\varepsilon$.
Question 1: Is my proof correct?
Question 2: Do you know another proof?
Remark: If $X$ is an infinite dimensional Banach space, then every $y\in Q$ belongs to the weak closure of $B_\varepsilon$.
Remark 1: Cross-Posted: mathoverflow