The claim should be that the intersection can be empty if this condition is dropped, not that it always will be.
Consider $X = \mathbb{N}$ with the discrete metric, and $C_n = \{n, n+1, n+2, \dots\}$.
A similar example would be $X=\mathbb{R}$ and $C_n = [n,+\infty)$.
Just because each of the sets $C_n$ contains at least two points (infinitely many, in these examples), does not mean that any one of those points has be common to all the sets, which is what would be necessary for the intersection to be nonempty.
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
This is false. For instance, let $X=\ell^\infty$ and let $C_n\subseteq X$ be the set of sequences whose first $n$ coordinates are all $0$. Then each $C_n$ has infinite diameter but their intersection is $\{0\}$.