A variation of the Riesz Lemma

Question

Given a normed space $X$, a closed proper subspace $Y$ and $\alpha\in (0,1)$, the Riesz Lemma states that there is $x\in X$ such that $\|x\|=1$ and $d(x,Y)>\alpha$. Observe that also $d(-x,Y)=d(x,Y)>\alpha$. That is, the distance from the unit sphere of $\langle x \rangle$ (where $\langle A\rangle$ denotes the linear span of $A$) to $Y$ is at least $\alpha$.

I'm looking for a proof or counterexample for the following variation of the Riesz Lemma:

Given a normed space $X$, a closed proper subspace $Y$ and $\alpha\in (0,1)$, there is $x\in X$ such that the distance from $\langle x\rangle$ to the unit sphere of $Y$ is at least $\alpha$.

I am mostly interested in positive answers, so partial positive answers would be nice if the result is false. Particularly, I'm only concerned with Banach spaces, so this can be assumed without problem.

EDIT: Pietro presented a clever construction in finite dimension and with $\mathrm{codim}(Y)=1$. Is there also a counterexample when $\dim(X)=\infty$ and $\mathrm{codim}(Y)=\infty$?

Answer 1

Let's denote $B$ the unit ball of $X$. Geometrically, we are asking if there is a line $\langle x\rangle$ such that the ball $\alpha B$ can be translated along $\langle x\rangle$ and pass through the hole of $Y\setminus B $, that is we seek a non-zero $x$ such that $$ \big( \alpha B+ \langle x\rangle\big) \cap \big(Y\setminus B\big)=\emptyset.$$

I think the simplest counterexample is: $X:=\mathbb R^3$ with the box norm $\|(a,b,c)\|_\infty:=\max(|a|,|b|,|c|\}$ and $Y:=(1,1,1)^\perp=\{(a,b,c):a+b+c=0\}$. The section $Y\cap B$ is a regular hexagon with vertices in mid-points of some edges, but every projection of $\alpha B$ onto $Y$ is a polygon with vertices in the projection of some vertices of $\alpha B$, thus never included in $Y\cap B$ for $\alpha$ too close to $1$. (Or just consider the angles between $Y$ and the faces of the cube).