Let$ ({F_j})_{j=1}^{\infty}$ be a sequence of bounded closed sets in $\mathbb{R}^d$ with the following properties.. Let$ ({F_j})_{j=1}^{\infty}$ be a sequence of bounded closed sets in $\mathbb{R}^d$ with the following properties..
$\cap^m_{j=1}$ $F_j$ is non empty for every $m \in\mathbb{N}$
Prove that $\cap^{\infty}_{j=1}$ $F_j$ is also non empty.
My thought process:
E is bounded if there exists $M>0$ such that $||x||\leq M$ for all x $\in$ E. So I was thinking that I could apply the Bolzano-Weierstrass theorem to the sequence $Xn_{n=1}^{\infty}$ for  $X_n$$\in$ $\cap^{n}_{j=1}$ $F_j$ for each n.
But I'm not sure how to do the actual proof?
 A: For every $j \in \mathbb{N}$ the set $F_j \cap F_1$ is closed in $F_1$ with the topology induced by $\mathbb{R}^d$. Moreover $F_1$ is compact with this topology.
Your hypothesis in particular tells you that every finite subfamily of $\{F_j \cap F_1\}_{j \in \mathbb{N}}$ has nonempty intersection. Then the intesection of the whole family, that is $\cap_{j=1}^\infty F_j$, is also nonempty.
Why? Well because this is a property that characterizes the compact spaces: a topological space $X$ is compact if and only if every family of closed sets of $X$ satisfying the so-called F.I.P. (that is, every finite subfamily of the given family has nonempty intersection) has nonempty intersection. It's not difficult to prove this characterization by using the usual definition of compact space and this can be a nice exercise.
A: Here is another.  For each $n\in \mathbb{N}$, choose $x_n \in F_n$.  This sequence has a convergent subsequence.  If this subsequential limit is $x$, it is not hard to see that $x\in \cap_nF_n$.
A: OK, since we've had a couple of false proofs (one of which has been deleted, as of writing), I'm going weigh in.
Let $G_n = \bigcap_{k=1}^n F_k$. Then $G_n$ is compact for all $n$. Further, $G_{n+1} \subseteq G_n$ for all $n$.
Let $x_n \in G_n$. Since $G_n \subseteq F_1$ for all $n$, we know that $(x_n)$ is a sequence in a bounded set $F_1$, hence it must contain a convergent subsequence $x_{n_k} \to x$.
As part of the definition of a subsequence, we have that $n_k$ is a strictly increasing sequence of natural numbers, and hence $n_k \ge k$ for all $k$. Therefore,
$$x_{n_k} \in G_{n_k} \subseteq G_k.$$
Therefore, for any $m$, the tail of this subsequence $(x_{n_k})_{k \ge m}$ lies in $G_m$, which is closed. Consequently, the limit of this subsequence $x$ must lie in $G_m$. This holds for all $m$, hence
$$x \in \bigcap_{n=1}^\infty G_n.$$
This, of course, implies
$$x \in \bigcap_{n=1}^\infty F_n.$$
A: I rephrase my proof so that it does not explicitly make use of the characterization of the notion of compact space based on the so-called $FIP$.
Again, observe that $F_1$ is a compact space with the topology induced by $\mathbb{R}^d$. Suppose by contradiction that: $$\bigcap_{j=2}^\infty (F_j\cap F_1)=\bigcap_{j=1}^\infty F_j=\emptyset.$$ Equivalently: $$F_1=\bigcup_{j=2}^\infty (F_j^c \cap F_1).$$ For every $j \in \{2,3,...\}$ it is the case that $F_j^c \cap F_1$ is an open set of $F_1$, as $F_j^c$ is an open set of $\mathbb{R}^d$. Being $F_1$ compact, there is $N \in \mathbb{N}$ such that: $$F_1=\bigcup_{j=2}^N (F_j^c \cap F_1)$$ that is: $$\bigcap_{j=1}^N F_j=\bigcap_{j=2}^N (F_j\cap F_1)=\emptyset.$$ This contradicts the hypothesis.
A: You are right Bolzano-Weierstrass works. Also, you can use that $\mathbb{R}$ is complete (overkill). I think that proof is in Rudin's book. The interesting part of that exercise is that it shows that $\mathbb{R}$ doesn't have "holes". 
