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 $H$ 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.