Conditional compactness iff totally bounded 
We say that a metric space $M$ is totally bounded if for every $\epsilon>0$, there exist $x_1,\ldots,x_n\in M$ such that $M=B_\epsilon(x_1)\cup\ldots\cup B_\epsilon(x_n)$.
A metric space in which every sequence has a Cauchy subsequence is said to be conditionally compact. Prove that a metric space $M$ is conditionally compact if and only if $M$ is totally bounded.

So suppose $M$ is totally bounded, and given a sequence $y_1,y_2,\ldots$ in $M$. I want to find a Cauchy subsequence. For any $\epsilon$, I can find $x_1,\ldots,x_n\in M$ such that $M=B_\epsilon(x_1)\cup\ldots\cup B_\epsilon(x_n)$. So some ball contains infinitely many points $y_i$. Those points are within $2\epsilon$ of each other. But this still doesn't give me a Cauchy subsequence, because as $\epsilon$ decreases, the balls can change...

The backward direction is proved in the comment. Here's the forward direction:
Suppose $M$ is conditionally compact. Fix $\epsilon$. Every sequence has a Cauchy subsequence. Choose $x_1\in M$. If $B_\epsilon(x_1)$ doesn't cover $M$, choose $x_2$ in $M-B_\epsilon(x_1)$. Note that $d(x_2,x_1)\ge\epsilon$. If $B_\epsilon(x_1)\cup B_\epsilon(x_2)$ doesn't cover $M$, choose $x_3$ in $M-B_\epsilon(x_1)-B_\epsilon(x_2)$. Note that $d(x_3,x_1),d(x_3,x_2)\ge\epsilon$. If this process doesn't terminate, we get a sequence $x_1,x_2,\ldots$ such that $d(x_i,x_j)\ge\epsilon$ for all $i,j$. This sequence clearly has no Cauchy subsequence.
