Nested sequence of closed sets

If $X$ is a compact space, this is true. $\mathcal{F} = \{I_\alpha\}_{\alpha \in I}$ is an arbitrary collection of nonempty closed sets such that any intersection of finitely many sets from $\mathcal{F}$ is nonempty, then $\bigcap_{\alpha \in I} I_\alpha$ is nonempty.

To prove this, suppose $\bigcap_{\alpha \in I} I_\alpha$ is empty. Then $\{X - I_\alpha\}$ is an open cover of $X$. Since $X$ is compact, there exists $\alpha_1, ..., \alpha_n$ such that $\bigcup_{j = 1}^n (X - I_{\alpha_j}) = X$. Then $\bigcap_{j = 1}^n I_{\alpha_j} = \emptyset$. This contradicts the assumption that any finite intersection of elements of $\mathcal{F}$ is nonempty.

For example (in your case) if $\mathcal{F}$ consists of nonempty nested intervals, then $\mathcal{F}$ has the finite intersection property and the above proof assured that the infinite intersection is nonempty.

No: take $I_n=[n,\to)=\{x\in\Bbb R:x\ge n\}$. If, however, some $I_m$ is compact, then $I_n$ is compact for all $n\ge m$, and the intersection will be non-empty.

I'm studying intro to analysis and I got that a nested sequence of closed and nonempty subsets of $\mathbb{R}$ do not necessarily have a nonempty infinite intersection:

$\textbf{Claim 1}$: The set $A_{n} = (-\infty, -n]$ is closed, $n\in \mathbb{N}$.

$\textit{proof}$: Let $x$ be a limit point for $A_{n}\Rightarrow$ if we had $x>-n,$ choose $\epsilon = x+n\Rightarrow x-\epsilon=-n<\displaystyle x-\frac{\epsilon}{2}\Rightarrow V_{\epsilon/2}(x)\cap A_{n}=\emptyset \Rightarrow $ contradiction $\Rightarrow x\in A_{n}$.

$\textbf{Claim 2}$: $A_{n+1}\subseteq A_{n}\forall n$.

$\textit{Proof}$: $x\in A_{n+1}\Rightarrow x\leq-n-1<-n\Rightarrow x\in A_{n}$.

$\textbf{Claim 3}$: The sequence $A_{n}\supset A_{n+1}, A_{n}=(-\infty,-n] \forall n$ has empty infinite intersection.

$\textit{Proof}: \bigcap_{n=1}^{\infty}A_{n}\neq \emptyset \Rightarrow$ choose $x\in \bigcap_{n=1}^{\infty}A_{n}\Rightarrow x\leq -n\hspace{.5pc}\forall n\Rightarrow -x\geq n\hspace{.5pc}\forall n\Rightarrow$ contradiction ($\mathbb{N}$ is not bounded above)

$\textbf{Conclusion}$: By claims 1,2, and 3, we have a nested sequence of closed sets with empty infinite intersection.

$\underline{Legend}$:

1. $V_{\epsilon}(x)=(x-\epsilon,x+\epsilon)$
2. Infinite intersection of $A_{n}=\bigcap_{n=1}^{\infty}A_{n}$