Here is a useful proof idea/outline for you.
Suppose, $\bigcap_{i=1}^{\infty}E_i = \emptyset$
Then choose $x_1\in E_1$ for instance. Then we know that $x_1\notin E_k$ fo some $k$, because otherwise the intersection would not be empty. So $x\in E_{k_1}^c$ for some $k_1$. We can do this for each $x_1\in E_1$ so we see that we can find an cover for $E_1$
This open cover is: $\{E_{k_i}^c\}$. Notice, although I wrote it like it is countable it need not be.
But since $E_1$ is compact we can find a finite subcover $E_{k_1}^c, \cdots, E_{k_n}^c$. (why can we do this? i.e. how do we know that we had an open cover?)
But then this means that $E_1 \subset \bigcup_{i=1}^{n}E_{k_i}^c = (\bigcap_{i=1}^n E_{k_i})^c$ (why?)
Which implies that $E_1\cap \bigcap_{i=1}^nE_{k_i} = \emptyset$.
However, by assumption we have that $E_k \supset E_{k+1}$ for all $k$: Use this fact to obtain a contradiction (Hint: Consider the assumption that the sets were non-empty)
Also, for the counterexample part. How about we replace "compact" by either closed or bounded and find counterexamples.
If we replace it with "closed": I suggest considering something which goes of to infinity
and "bounded": I suggest considering something which does not contain its limit points (i.e is not closed)