Suppose $x,x_n\in c_0$ with norm 1 such that $\|x_n+x\|\to 2$. Let $f\in c_0^*$ with $\|f\|=1$ such that $f(x)=1$. I want to check if $$d(x_n,A)\to 0, \text{ where }A=\{y\in c_0: \|y\|=1, f(y)=1\}$$
I tried to find a counterexample to disprove it, but could not do so. Neither can I prove it. Any suggestion please!
Counterexample: \begin{align*} x &= (1, 1, 0, 0, \ldots) \\ x_n &= \begin{cases}(1, 0, 0, 0, \ldots) & \text{if } n \text{ is even} \\ (0, 1, 0, 0, \ldots) & \text{if } n \text{ is odd}.\end{cases} \end{align*} Note that $\|x + x_n\| = 2$.
Further, let $f(y)$ be the first coordinate of $y \in c_0$. Note that $f \in c_0^*$. It corresponds to the sequence $(1, 0, 0, \ldots)$ in $\ell^1$, and has norm $1$. We also have $f(x) = 1$.
However, $f(x + x_n)$ oscillates between $1$ and $2$. In particular, if $n$ is odd, then $f(x_n) = 0$, and if $y \in A$, then $\|y - x_n\|$ is at least $1$. Thus, $d(x_n, A)$ does not tend to $0$.