Can the intersection of open or closed balls be empty, if their radii are bounded from below? I am wondering about the following question: 
Given a (countable) sequence of nested open balls: 
$$ B_1 \supseteq B_2 \supseteq \cdots $$
Not necessarily having the same same center. All having radius bounded from below, say by $r > 0$. Then can we say that $$\bigcap_{i=1}^{\infty} B_i \neq \varnothing$$
it is certainly true in the reals, as one can simply go to the point where the radii are close to $r$ then take the center of that ball. However, I'm having trouble seeing whether or not it is true in the general case. So thanks in advance for proof or counterexample. 
 A: Let $N$ be the set of positive integers. Define a metric on $N$ as follows:
$$d(m,n)=\left\{\begin{array}{ll} 1+ \frac{1}{mn}& \mbox{if }m \neq n\\ 0 &\mbox{if }m=n \end{array}\right.$$
It is straightforward to check that this is a complete metric on $N$ (complete because this metric is discrete). 
The closed balls  $$B(n, 1+1/n^2)=\{m: m\ge n\}, n=1,2,\cdots$$ are decreasing and have
empty intersection. Of course, the corresponding open balls also have empty intersection.
A: The below argument is incorrect as pointed out by TCL. I'll leave the answer here anyway, just incase.

Note that $\bigcap_i^n B_i=B_n$ because each $B_i$ is a subset of $B_{i+1}$. Let $d_i=\mbox{diam}(B_i)$. It should be clear that if $\bigcap_i^{\infty} B_i=\emptyset$, then $\lim_{i\rightarrow\infty}d_i=0$ and so there exists some $i\in\mathbb{N}$ such that $d_i< 2r$, but then $B_i$ is an open ball of radius less than $r$ which contradicts the assumption that each $B_i$ has radius bounded below by some non-zero $r$.
I should add that i'm using the definition of the diameter of a subset $A$ of a metric space $X$ with metric $d$ to be $\mbox{diam}(A)=\sup\{d(x,y)|x,y\in A\}$.
