Topology: Prove the intersection of these sets is non empty. The question is as follows:  


  Let $C_n$ be a sequence of nonempty, closed subsets of a compact
  space $X$, such that, $C_n ⊃ C_{n+1}$ for each $n$. Show that the intersection
  of these sets is nonempty.   


My question is where in my proof(below) am I wrong as I do not explicitly state that it is a compact space.
The unions below are infinite. 

$Attempt:$

$C_0 \neq \emptyset $
$(X-C_0)\cup(\cup(X-C_i)) = X - \cap C_i$
$C_0$ is non empty and then $X-C_i$ is not X and is either a subset of X or the null set.
If it is not the null set then their difference is non empty and not X.
So:
$X-\cap C_i= A$, where A is a non empty subset of X.
Then the complement is non empty and thus our intersection is non empty.

If it's the null set then we reapply our original procedure for a subset in the sequence which is greater than the null set and arrive at the same result.
You're thinking along the right lines, but your proof needs to use the definition compactness somewhere, since it's not true without that assumption.

Here's a (big) hint. If it were the case that $\bigcap\limits_{n=0}^{\infty} C_n = \varnothing$, then we'd have
$$X = X - \varnothing = X - \bigcap_{n=0}^{\infty} C_n = \bigcup_{n=0}^{\infty} (X-C_n)$$
and so $\{ X-C_n \mid n \ge 0 \}$ is an open cover of $X$. By compactness, it has a finite subcover and so, in particular, there is some $k \in \mathbb{N}$ such that $\{ X-C_n \mid 0 \le n \le k \}$ is a cover of $X$.

Use this observation to derive a contradiction to one of the hypotheses about the sets $C_n$.
The intersection of $C_i$ with $C_{i+1}$ is $C_{i+1}$, therefore the intersection of all of the sets is $C_l$ where $l$ is either the maximum value of $n$ or the limit as $n$ goes to infinity. Since all $C_n$ are non-empty, $C_l$ is also non-empty and the intersection of all $C_n$ is also non-empty.