Note: The following examples show that the conditions $\displaystyle \lim_{x\to\infty} d(F_{n})=0$ and that $F_{n}$ are closed sets both are necessary for the validity of the theorem.
Example: Let $X$ be the real line $R$ and let $F_{n}=[n,\infty).$ Now we know that $X$ is complete, $F_{1}\supset F_{2}\supset F_{3}....$ and $F_{n}$ are closed sets. But $\displaystyle \bigcap_{n=1}^{\infty} F_{n}=\phi.$Note that $\displaystyle \lim_{n\to\infty}d(F_{n})\neq 0.$
Example: Let $X$ be the real line $R$ and let $\displaystyle F_{n}= ( 0, \frac{1}{n}].$ Now we know that $X$ is complete, $F_{1}\supset F_{2}\supset F_{3}....$ and $\displaystyle \lim_{n\to\infty}d(F_{n})=0.$ But $\displaystyle \bigcap_{n=1}^{\infty} F_{n}=\phi.$ Note that the $F_{n}'s$ are not closed.
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:
Claim: Suppose that $$A_1\supseteq A_2\supseteq A_3\supseteq\ldots$$ is a countable family of non-empty, closed, bounded subsets of $\mathbb R$. Then, $\bigcap_{n=1}^{\infty} A_n\neq\varnothing$.
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. $\quad\blacksquare$
Best Answer
Consider the intersection of all sets of the form $[n,\infty)$, where $n$ ranges over the positive integers.