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:



> 
> Question 1: Can we find $y\in X$ such that $\|y\|<1$ and any line which contains $y$ must intersect $B\_\varepsilon$?
> 
> 
> 


Now consider the following solution in the case where $X$ is finite dimensional:


Take a [supporting hyperplane](https://en.wikipedia.org/wiki/Supporting_hyperplane) $H$ for the convex set $B=\{x\in X:\|x\|\le 1\}$ at the point $x\_0$. 



> 
> Claim: There exists $z\in X\setminus\{0\}$ such the set
>  $$
> H\_z=\{h+z:h\in H\},
> $$
>  has the property that $H\_z\cap B\_\varepsilon\neq \emptyset$ and $H\_z\cap B\_\varepsilon=H\_z\cap S$. 
> 
> 
> 


Indeed, we claim that there exists $\delta>0$ such that $z=-\delta x\_0$. Since $H$ is a supporting hyperplane it follows that if $\delta$ is small, then $H\_{-\delta x\_0}\cap B\_\varepsilon\neq \emptyset$. Now suppose on the contrary that there exists a sequence $\delta\_n>0$ such that $\delta\_n\downarrow 0$ and a corresponding sequence $x\_n\in H\_{-\delta\_n x\_0}\cap S$ satisfying $x\_n \notin B\_\varepsilon$. 


Once $x\_n$ is bounded, we can assume without loss of generality that $x\_n\to x\in H\cap S$. If $x=x\_0$ we get a contradiction since $x\_n\in B\_\varepsilon$ for bigger $n$. If $x\neq x\_0$, we also get a contradiction with the strictly convexity, by observing that the segment $\{tx\_0+(1-t)x:\ t\in [0,1]\}$ should belong to $B\_\varepsilon$. Therefore the Claim is true.


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 2: What about the infinite dimensional case?
> 
> 
> 


Remark 1: If $X$ is an infinite dimensional Banach space, then every $y$ which solves Question 1 belongs to the weak closure of $B\_\varepsilon$. 


Remark 2: Cross-posted: [math.stackexchange](https://math.stackexchange.com/questions/3025716/weak-closure-of-subsets-of-the-unitary-sphere-of-a-banach-space).



Weak closure of subsets of the unitary sphere of a Banach space
If you look at the definition of a locally uniformly convex Banach space in, for example, my book Ostrovskii, Metric Embeddings, page 291, you will see that any point $x\_0$ at which the condition of this definition is satisfied, has the property described in your Question 1. The converse does not seem to be true, as it seems that you need more than the negation of local uniform convexity, you need a kind of symmetric lack of the local uniform convexity.


As an example of a strictly convex Banach space without the property described in Question 1 I suggest the space $c\_0$ with the norm $$||\{a\_i\}\_{i=1}^\infty||=\max\_i|a\_i|+\left(\sum\_{i=1}^\infty |a\_i|^2/2^i\right)^{1/2}.$$


