If $X$ satisfies the decreasing chain condition then $X$ is compact? Let $X$ be a metric space. We say that $X$ satisfies Condition $C•$ if and only if, for
any decreasing sequence:
$$
··· ⊆ C_j ⊆ ··· ⊆ C_3 ⊆ C_2 ⊆ C_1
$$
of nonempty closed subsets of $X$, the intersection:
$$
\bigcap_{j=1}^\infty C_j 
$$
is nonempty. Show that if $X$ satisfies Condition $C•$ then $X$ is compact.
So my understanding is that this question if closely related to Cantor's Intersection Theorem which says that if $C•$ condition and compact then the intersection is non empty. But I am struggling to show that if condition $C•$ is met then $X$ has to be compact. 
I tried proceeding by way of contradiction (suppose that $X$ is not compact which means that $X$ is not complete, not totally bounded (remember that this is in a metric space so compactness is complete and totally bounded and not just closed and bounded)). Any help would be appreciated. 
 A: Using your condition that $X$ is compact if and only if $X$ is complete and totally bounded. Assume that $X$ is not compact, then $X$ is either not complete or not totally bounded. 
If $X$ is not complete, there is a Cauchy sequence $\{x_n \}$ so that it has no limit. Let $C_n = \{ x_k: k\ge n\}$. Then $C_n$ are all closed (if $C_n$ has a limit point $x$, then a subsequence of $\{x_n\}$ converges to $x$, this forces $\{x_n\}$ to converge to $x$ as $\{x_n\}$ is Cauchy). However, 
$$\bigcap _n C_n = \emptyset.$$
Thus $X$ does not satisfies condition $C•$. 
If $X$ is not totally bounded, then there is $\epsilon >0$ and $\{x_i\}_{i\in I}$ so that $X$ is not covered by $\bigcup_{j\in J} B_\epsilon(x_j)$ for any finite subset $J\subset I$. Pick $i_1 \in I$ and call $y_1 = x_{i_1}$. Then there is $i_2 \in I$ and $y_2 \in B_\epsilon(x_{i_2})$ so that $y_2 \notin B_\epsilon (x_{i_1})$. 
Assume inductively that you have chosen $y_k \in B_\epsilon (x_{i_k})$, $k=1, \cdots, n$ so that $y_k \notin \bigcup_{ j\le k} B_\epsilon (x_{i_j})$ for all $k = 2, \cdots, n$. Then as $\bigcup_ k B_\epsilon(x_{i_k})$ does not cover $X$, there is $i_{n+1}$ and $y_{n+1} \in B_\epsilon (x_{i_{n+1}})$ so that $y_{n+1} \notin \cup_k\le n B_\epsilon (x_{i_k})$. 
Thus we have found a sequence $\{y_n\}$ in $X$ so that $d(y_i, y_j) \ge \epsilon$ for all $i, j$. Thus again $C_n = \{y_k : k\ge n\}$ is closed and violate condition $C•$. 
