$I=\cap B_i$ is closed because the arbitrary intersection of closed sets is closed hence $\cap B_i$ is closed. As you mentioned, any compact subset of $\mathbb{R}^n$ is closed and bounded
Now $I$ is a closed subset of a compact set (any $B_i$ will suffice). Hence it is also compact (Why? Proof below). To answer your question about the boundedness of $I$, it is a subset of a bounded set (any $B_{i}$ suffices) hence it is bounded.
We need to prove that a closed subset of a compact space is closed.
Let $X$ be any hausdorff space. ($\mathbb{R}^n$ certainly is.)
Let $A\subset C$ be closed with $C$ compact. Take an open covering of $M=\{A_{\alpha}\}$. Since $A$ is closed, $U=X-A$ is open. Hence $N= M\cup \{U\}$ is an open covering of $C$. Since $C$ is compact, $M$ has a finite subcover for $C$. This obviously covers $A$ as well thus $A$ has a finite subcover. If the finite subcover of $C$ involves $U$, we can always eliminate it.
I assumed $X$ is hausdorff because it can be shown that any compact subset of a hausdorff space is closed.
Is it sufficient to show that ${\bigcap}_{i=0}^{\infty} A_i$ is bounded and closed ?
No, this is not sufficient. There exist sets which are bounded and closed, yet they are not compact. For example, the set $(0,1)$ is abounded closed subset of the space $(0,1)$, yet the set is not compact.
There are two ways I see that you can solve the question:
Option 1:
There is a theorem that states that a closed subset of a compact set is compact. This means all you need to prove is that the intersection in question is closed, and since it is a subset of $A_1$, a compact set, you are done.
Option 2:
If you cannot yet use the above theorem (this is probably the case, since otherwise, the problem is way too easy), you can just go from definitions. That is, start with an open cover for ${\bigcap}_{i=0}^{\infty} A_i$, and try to find some finite subcover. In order to do that, think about this fact:
Any open cover of ${\bigcap}_{i=0}^{\infty} A_i$, along with the set $X\setminus {\bigcap}_{i=0}^{\infty} A_i$, is an open cover of $A_1$...
Best Answer
The proof is somewhat dependent on the definitions you use.
The simple (as you basically have only one possible definition) part is the intersection of bounded sets. You either define bounded as having a radius such that every member of the set has absolute value (or norm) less or equal to the radius, or more general that the distance of any two points in the set have distance $d(x,y)$ less or equal to the radius (I use the latter), I don't require the radius to be the smallest bound here (if one bound exists there exists an infimum of them which is also a bound by the axiom of largest lower bound):
Let $B_j$ be bounded sets which means that there exists a radius $r_j$ such that $d(x,y)\le r_j$ for all $x,y\in B_j$. Now consider the intersection $B = \bigcap B_j$, then we have that $d(x,y)\le r_o$ for any $x,y\in B_o$, but since $B \subseteq B_o$ we have for any $x, y\in B$ that $x,y\in B_o$ and therefore $d(x,y)\le r_o$, so $r_o$ is a radiuos of $B$.
To prove that intersection of closed sets is closed we have to do that differently depending on the definition:
Also for compactness it's a matter of definition: