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.
 A: 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 $C_n$.
A: Counterexample: $C_n=[n,\infty)$ in $\mathbb{R}$ with the usual topology. The sets are closed and $C_{n+1}\subset C_n$ for every $n$. However the intersection is empty.

Definition. A family $\mathcal{A}$ of subsets of $X$ has the finite intersection property if every finite family of members of $\mathcal{A}$ has non empty intersection.
Theorem. A topological space $X$ is compact if and only if every family $\mathcal{A}$ of closed subsets of $X$ having the finite intersection property has non empty intersection.
Proof. Suppose $X$ is compact and that $\mathcal{A}$ is a family of closed subsets of $X$ having empty intersection. Then $\mathcal{U}=\{X\setminus A:A\in\mathcal{A}\}$ is an open cover of $X$ by De Morgan's laws:
$$
\bigcup\mathcal{U}=\bigcup_{A\in\mathcal{A}}(X\setminus A)=
X\setminus\Bigl(\,\bigcap_{A\in\mathcal{A}}A\Bigr)=X
$$
Then there exist $A_1,A_2,\dots,A_n\in\mathcal{A}$ such that
$$
X=(X\setminus A_1)\cup\dots\cup(X\setminus A_n)=
X\setminus(A_1\cap\dots\cap A_n)
$$
implying $A_1\cap\dots\cap A_n=\emptyset$, so $\mathcal{A}$ doesn't have the finite intersection property.
For the converse, suppose you're given an open cover $\mathcal{U}$ of $X$. Then the family $\mathcal{A}=\{X\setminus U:U\in\mathcal{U}\}$ is a family of closed sets with empty intersection, again by De Morgan's laws. Then $\mathcal{A}$ doesn't have the finite intersection property, so there is a finite subfamily having empty intersection, which provides the required subcover of $\mathcal{U}$.$\quad□$
Obviously, a decreasing family of nonempty subsets has the finite intersection property.
A: If you are using limit point compact, try choosing a sequence $x_i\in C_i$.
The set of points of the sequence cannot be finite unless at least one of the points lies in $\cap C_i$. If it is infinite then some subsequence converges to a point $P$.
Can you show that $P$ must lie in $\cap C_i$?
A: 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.
