Metric spaces in topology $(X,d)$ be a metric space. Given $\epsilon >0$ there is a non empty finite subset $X_\epsilon\subset X$ such that for every $x\in X$, we have inf$\{d(x,p):p\in X_\epsilon\}\leq\epsilon$. 
1) Show that in this case, every sequence in $X$ has a cauchy subsequence.
2) Give an example showing that (1) fails if we do not require the $X_\epsilon$'s to be finite.
I proved that $X$ is totally bounded. If I can prove $X$ is complete, then it completes part (1). How can I prove $X$ is complete? And also need an example for part (2).
 A: Take any sequence $x_n$, $n=1,2...$ Take $\epsilon=1/2$. There are finitely many open balls of radius $1/2$ that cover $X$ (i.e. their union is $X$). Hence there is an infinite subsequence that is contained in just one of these balls, let $y_1$ be the first element of this subsequence. Call this subsequence "the first subsequence". Now take $\epsilon=1/4$. There are finitely many open balls of radius $1/4$ that cover $X$. Hence "the first subsequence" has a subsequence all elements of which are in just one of these balls of radius $1/4$. Call this latter subsequence "the second subsequence" and pick one of its elements $y_2$ (and such that $y_2$ comes later than $y_1$ in the original sequence). Next take $\epsilon=1/8$ and use it to find a "third subsequence" and pick $y_3$ from it (that comes later than $y_2$ in the original sequence). Proceeding this way, define a subsequence $y_1,y_2,y_3,...$ of the sequence $x_n$ such that $d(y_n,y_{n+1})<1/2^{n-1}$ (since both $y_n$ and $y_{n+1}$ are contained in the same ball of radius $1/2^n$). To show that the $y_n$ form a Cauchy sequence, take any $\varepsilon>0$ and take $N$ such that $2^n<\varepsilon/4$. If $k,m\ge N$ then $d(y_k,y_m)<\sum_{n\ge N}d(y_n,y_{n+1})<1/2^{n-2}<\varepsilon$. 
