Intersection of closed and bounded subsets contained within open set $\textbf{Problem statement}:$
Suppose that $\{F_j\}$ are closed and bounded subsets of $\mathbb{R}$, and $G$ is an open subset. Show that if 
$
\cap_{j=1}^{\infty} F_j \subset G, 
$
then there is a finite subset $\{j_1,\dots,j_k \}$ such that $\cap_{i=1}^k F_{j_i} \subset G$. 
Let the open set $G$ be $(a,b)$, and I have showed that the countable intersection of closed sets is closed already, and since each $F_j$ is bounded, the total intersection is some $[a',b'] \subset (a,b)$. 
Now assume that there is no such finite subset $\{j_1,\dots,j_k \}$ such that $\cap_{i=1}^k F_{j_i} \subset G$. We know that for each finite intersection of closed and bounded sets, the intersection is closed and bounded, and the lower bound is $sup(inf(F_j))$ which converges to $a'$, and the upper bound is $inf(sup(F_j))$ which converges to $b'$. Since we have this convergence, $\vert sup(inf(F_j)) - a'\vert < \epsilon$ and $\vert inf(sup(F_j)) - b' \vert < \epsilon$. Then we can find an $N_1$ such that $\vert sup(inf(F_{N_1}) - a'\vert < \frac{a' - a}{2}$ equal to $min(\frac{a' - a}{2},\frac{b - b'}{2})$, $N_2$ so $\vert inf(sup(F_{N_2})) - b' \vert < \frac{b - b'}{2})$ and take $N = max(N_1,N_2)$ which is a finite $N$ such that the intersection of those $N$ subsets is contained in $(a,b)$ and thus a contradiction. 
I doubt the rigor of this proof to be honest, and feels messy. If anyone can correct me or provide something cleaner I'd appreciate it. A hint on generalization to $\mathbb{R}^n$ would also be helpful. 
 A: Your proof assumes that the intersection is a segment in $\mathbb{R}$, which does not have to be true.

Assume there does not exist a finite $I \subseteq \mathbb{N}$ such that $\bigcap_{i \in I} F_i \subseteq G$.
Then in particular, $F_1 \not\subseteq G$ so there exists $x_1 \in F_1 \setminus G$.
Also, $F_1 \cap F_2 \not\subseteq G$ so there exists $x_2 \in (F_1 \cap F_2 )\setminus G$.
Inductively, we arrive at a sequence $(x_n)_{n=1}^\infty$ such that $x_n \in \left(\bigcap_{i=1}^n F_i\right) \setminus G$ for all $n \in \mathbb{N}$.
Notice that $(x_n)_{n=1}^\infty$ is a sequence in $F_1 \setminus G$, which is closed and bounded and hence compact. Therefore, there exists a convergent subsequence of $(x_n)_{n=1}^\infty$ with the limit in $F_1 \setminus G$. WLOG we can assume that $x_n \xrightarrow{n\to\infty } x \in F_1 \setminus G$.
Note that for any $k \in \mathbb{N}$ we have that $(x_n)_{n \ge k}^\infty$ is a sequence in $\left(\bigcap_{i=1}^k F_i\right) \setminus G$ which is a closed set so its limit $x \in \left(\bigcap_{i=1}^k F_i\right)\setminus G$.
Thus, $x \in \left(\bigcap_{i=1}^\infty F_i\right)\setminus G$, which is a contradiction with the assumption that $\bigcap_{i=1}^\infty F_i \subseteq G$.
