I tried prove that if $F_1 \supseteq F_2 \supseteq F_3 \supseteq \dots $ is a nested sequence of nonempty closed sets then the intersection $\bigcap_n F_n \neq \varnothing$. I don't know for sure that it is true. Please can you check my proof?
Proof: If $F_i$ are all unbounded then the intersection is also unbounded and therefore not empty. If there is $i$ with $F_i$ bounded then all $F_n$ with $n>i$ are also bounded and therefore compact. Because the intersection of non-empty compact nested sets is nonempty in this case the intersection is also nonempty and therefore the statement is true.
Best Answer
$F_{n}:=\left[n,\infty\right)\subset\mathbb{R}$ is closed with $F_{1}\supset F_{2}\supset F_{3}\supset\cdots$
However $\cap_{n}F_{n}=\emptyset$.
Under extra condition that the $F_{n}$ are bounded your statement is true. Then you are dealing with compact sets.