Is Cantor's Intersection theorem valid for only a countable number of nested closed sets? Cantor's Intersection theorem says if $F_{n+1}\subset F_n$ $\forall n\in\Bbb{N}$, then $\bigcap_{i=1}^{\infty}F_i\neq\emptyset$. 
This is valid only in complete metric spaces, and the proof is based on a cauchy sequence formed by choosing one $x_k$ from each $F_k$. The number of terms in a convergent sequence is countable, and hence, it makes sense that the number of nested closed sets $F_i$ is also countable. 
Is there a way to generalise this to an uncountable number of nested closed sets? Say $F_m\subset F_n\forall m,n\in\Bbb{R},n>m$? If this were possible, it woud prove $\Bbb{R}$ is not the union of an uncountable number of nowhere dense sets, which would be a contradiction. 
Thanks in advance!
 A: Let $I$ be an uncountable index set endowed with a well ordering $\succeq$ (this is possible by the axiom of choice). For any $i\in I$, let $F_i$ be a nonempty, bounded, and closed subset of a complete and totally bounded metric space $S$ (hence, $F_i$ is compact for any $i\in I$). Moreover, suppose that $F_i\subseteq F_j$ whenever $i,j\in I$ such that $i\succeq j$. I will show that $F\equiv\bigcap_{i\in I} F_i$ is nonempty.
Suppose that $F$ is empty. Let $V_i\equiv S\bigcap F_i^c$ for any $i\in I$. Then, 
\begin{align*}
\bigcup_{i\in I}V_i=S\bigcap\left(\bigcup_{i\in I} F_i^c\right)=S\bigcap\left(\bigcap_{i\in I}F_i\right)^c=S\bigcap\varnothing^c=S.
\end{align*}
Therefore, $\{V_i\}_{i\in I}$ is an open cover of $S$ and hence also of any subset of $S$.
Let $k$ be the least element in $I$ according to the well order $\succeq$. Then, $F_i\subseteq F_k$ for any $i\in I$. Since $F_k$ is compact and $\{V_i\}_{i\in I}$ is an open cover of it, there is an open subcover $\{V_i\}_{i\in J}$, where $J\subset I$ is a nonempty finite set. Let $\ell$ be the greatest element in $J$. Then, $\bigcup_{i\in J}V_j=V_{\ell}$ by construction. Hence, $F_k\subseteq V_{\ell}$, which implies that $F_i\subseteq V_{\ell}$ for any $i\in I$. But this is a contradiction, since $F_{\ell}$ is nonempty (so that $x\in F_{\ell}$ should imply that $x\in V_{\ell}\subseteq F_{\ell}^c$).
Therefore, $F\neq\varnothing.$
A: Suppose the $F_k$ are closed and bounded, for each real $k\ge 1$, and that $F_m \subset F_n$ whenever $m > n.$ Then the subsequence $F_1,F_2,F_3,\cdots$ satisfies the hypothesis of Cantor's intersection theorem, so that the intersection of these $F_k$ is nonempty, so contains some specific point $z.$ But then the intersection of all the $F_m$, $m \in [1,\infty)$ is also nonempty, since for any $m$ we can choose a $k>m$ with $k$ an integer, so that $F_k \subset F_m$ and thus $z \in F_m.$
