If $F$ decreasing closed sequence and $F_1$ is bounded then
$\cap_{n=1}^{\infty}F_n\not= \emptyset$
Attempt at a proof: Let assume that that $\cap_{n=1}^{\infty}F_n= \emptyset$ (to arrive at a contradiction) so $\forall \not{\exists}n: x_n\in F_n$
but $F_{n+1}\subset F_n \subset \dots \subset F_1 $
How to continue? must $F_1$ be bounded?
This is true only if the closed sets are compact. Pick, for example, the sets $F_n=[n,+\infty)$. They are all closed sets, the sequence is decreasing, but $\cap F_n=\emptyset$. By this counterexample, we get that $F_1$ must be bounded.
Conversely, if $F_1$ is bounded and closed, then as a property of $\mathbb R^n$, $F_1$ is a compact set, so it follows that it has the finite intersection property.