Let's ignore the ball condition at first, and just look for a complete metric space and a nested sequence of nonempty closed sets with empty intersection. This is clearly impossible for a finite metric space, so let's try the simplest possible infinite metric space: a countable set, which might as well be $\mathbb{N}$, with the discrete metric $d(m,n)=1 \Leftrightarrow m \neq n$. Every set in this space is closed, so it's easy to find the sequence we want: for example, we can take $A_n$ to be $\{ n, n+1, n+2, \ldots\}$.
This is all very well, but of course these sets aren't actually closed balls yet. This doesn't mean we should give up on our example, though. Rather, we try to tweak the metric to make each $A_n$ into a closed ball. We also need our metric to stay complete, and the simplest way to ensure this is to keep the condition that all nonzero distances are $\ge 1$.
A ball is defined by two things: a centre and a radius. The simplest choice for centre is to have $A_n$ centred at $n$. The radii will have to be decreasing (since we want $n+1 \in A_n$ but not $n \in A_{n+1}$) as well as being at least $1$, so let's try to make the radius of $A_n$ be $1+\frac1n$.
So for $m<n$, we want $d(m,n)\le 1+\frac1m$ (so that $n$ is in $A_m$) but $d(m,n)>1+\frac1n$ (so that $m$ is not in $A_n$). The simplest way to do this is to just let $d(m,n)=1+\frac1m$.
All the above is just a way of motivating the following example: let $X$ be the space $\mathbb N$ with metric $d(m,n)=1+\frac{1}{\min (m,n)}$ for $m \neq n$. It's not hard to check that this is indeed a complete metric space, and the closed ball with centre $n$ and radius $1+\frac1n$ is $\{n, n+1, n+2, \ldots\}$.
Best Answer
Let $\mathbb N$ be endowed with the discrete metric. In this metric space, every subset is bounded (although not necessarily totally bounded) and closed. Moreover, the subsets \begin{align*} A_1\equiv&\,\{1,2,3,4,\ldots\},\\ A_2\equiv&\,\{\phantom{1,\,}2,3,4,\ldots\},\\ A_3\equiv&\,\{\phantom{1,2,\,}3,4,\ldots\},\\ \vdots&\, \end{align*} are nested, and their intersection is empty.
However, if you stay within the realm of $\mathbb R$ endowed with the usual Euclidean metric, than you can't have a situation like the one above:
Proof: By the Heine–Borel theorem, $A_n$ is compact for each $n\in\mathbb N$. For the sake of contradiction, suppose that $\bigcap_{n=1}^{\infty} A_n=\varnothing$. This is equivalent to $\bigcup_{n=1}^{\infty} A_n^{\mathsf c}=\mathbb R$. In particular, $$A_1\subseteq\bigcup_{n=1}^{\infty} A_n^{\mathsf c}.$$ Since $A_1$ is compact and the sets $(A_n^{\mathsf c})_{n=1}^{\infty}$ form an open cover of it, there must exist a finite subcover. That is, there exists some $m\in\mathbb N$ such that $$A_1\subseteq\bigcup_{n=1}^m A_n^{\mathsf c}=A_m^{\mathsf c},$$ where the second equality follows from the fact that $$A_1^{\mathsf c}\subseteq A_2^{\mathsf c}\subseteq A_3^{\mathsf c}\subseteq\ldots.$$ Now, $A_1\subseteq A_m^{\mathsf c}$ means that if a point is in $A_1$, then it must not be in $A_m$, so that $A_1\cap A_m=\varnothing$. But $A_m\subseteq A_1$ (given that the sets are nested), so that $A_1\cap A_m= A_m=\varnothing$, which contradicts the assumption that $A_m$ is not empty. This contradiction reveals that the intersection $\bigcap_{n=1}^{\infty} A_n$ must not be empty. $\quad\blacksquare$