Nested families of balls in Banach spaces Is it possible to find a Banach space $X$, $x_i\in X$ and $r_i > 0$ indexed by some $i\in I$, possibly infinite, such that for each $n\in \mathbb{N}$
$$\bigcap_{i\in I} B\big(x_i, (1+\tfrac{1}{n})r_i\big)\neq \varnothing$$
but
$$\bigcap_{i\in I} B\big(x_i, r_i\big)= \varnothing ?$$
Here $B(x,r)$ stands for the closed ball centred at $x$ with radius $r$.
There is no restriction on cardinality of $I$ here (for instance, $I$ may be uncountable). Also the radii $r_i$ may be unbounded too.
Note that $X$ cannot be isometric to a dual space (which rules out $\ell_1$ and all reflexive spaces, for example). For those who don't believe, here is the proof. If $X$ is a dual space, then all closed balls are weakly* compact by the Banach-Alaoglu theorem. However, 
$$\bigcap_{i\in I} B\big(x_i, r_i\big) = \bigcap_n \bigcap_{i\in I} B\big(x_i, (1+\tfrac{1}{n})r_i\big) $$
must be non-empty being the intersection of a descending sequence of compact sets.
 A: I think I have found an example in $C[-1,1]$ with the usual maximum-norm.
We set $I := \{2, 3, \ldots\}$. We define the functions
\begin{equation*}
 x_i(t) :=
 \begin{cases}
  -i & \text{if } t \in [-1,-1/i), \\
  i^2 \, t & \text{if } t \in [-1/i, 1/i), \\
  i & \text{if } t \in [1/i,1],
 \end{cases}
\end{equation*}
and set $r_i := i-1$.
The emptiness of $\bigcap_{i \in I} B(x_i, r_i)$ is easy to establish.
Indeed, any element $y$ in this intersection would satisfy
$y(t) \le -1$ for $t < 0$ and $y(t) \ge 1$ for $t > 0$
(take $i > 1/t$).
Now, we define
\begin{equation*}
 y_n(t) :=
 \begin{cases}
  -1 & \text{if } t \in [-1,-1/n), \\
  n \, t & \text{if } t \in [-1/n, 1/n), \\
  1 & \text{if } t \in [1/n,1],
 \end{cases}
\end{equation*}
and show $y_n \in \bigcap_{i \in I} B(x_i, (1+1/n) \, r_i)$.
For $i \ge n$ we have $(1+1/n) \, r_i = (n+1) \, (i-1) / n \ge i$,
thus $\|x_i - y_n\| \le i \le (1+1/n)\,r_i$.
It remains to study $i < n$. This is easy via a distinction of cases:
\begin{align*}
 t \in [0,\frac1n]: &\quad | x_i(t) - y_n(t) | = |i^2 \, t - n \, t| \le |i^2-n|/n \le i-1, \\
 t \in [\frac1n,\frac1i]: &\quad | x_i(t) - y_n(t) | = |i^2 \, t - 1| \le i-1, \\
 t \in [\frac1i,1]: &\quad | x_i(t) - y_n(t) | = i - 1.
\end{align*}
The cases with $t < 0$ follow by symmetry.
Hence,
\begin{equation*}
 y_n \in \bigcap_{i \in I} B(x_i, (1+1/n) \, r_i).
\end{equation*}
