Let $(\mathbb{R}^n,\lVert\cdot\rVert)$ be a real, finite dimensional Banach space. Assume $B^\prime:=\{x\in \mathbb{R}^n\colon\langle x,y\rangle\leq 1\text{ for any }y\in B_1(0)\}$, where $B_1(0)$ is the unit ball in $\mathbb{R}^n$ with respect to the norm $\lVert\cdot\rVert$.
Let $x,y\in\mathbb{R}^n$ and $u,v\in B^\prime$ be such that $\langle x,u\rangle=\lVert x\rVert$ and $\langle y,v\rangle=\lVert y\rVert$. Is it true that the there is a vector $w\in\text{span}(u,v)\cap B^\prime$ such that $\langle w,y-x\rangle=\lVert x-y\rVert$?
A reference will be really appreciated.
No, this is too much to hope for. The duality map (which sends every element $x$ of a normed space to a unit-norm linear functional that evaluates to $\|x\|$ on $x$) is highly nonlinear in general. For example, let's take $\mathbb{R}^3$ with the norm $\|x\|_4 =(x_1^4+x_2^4+x_3^4)^{1/4} $. The duality map sends $x$ to $$ \phi(x) = \frac{1}{\|x\|_4^3}(x_1^3, x_2^3, x_3^3) $$ as one can check by observing that the $\ell^{4/3}$ norm of $\phi(x)$ is $1$, and that $\langle \phi(x), x\rangle = \|x\|_4$. Now, $$ \phi((1, 2, 3)) = \frac{1}{\|(1, 2, 3)\|_4^3}(1, 8, 27) $$ $$ \phi((1, 1, 1)) = \frac{1}{\|(1, 1, 1)\|_4^3}(1, 1, 1) $$ but for the difference $(1, 2, 3) - (1, 1, 1) = (0, 1, 2)$ we get $$ \phi((0, 1, 2)) = \frac{1}{\|(0, 1, 2)\|_4^3}(0, 1, 8) $$ which is not in the linar span of $(1, 8, 27)$ and $(1, 1, 1)$.