I'm trying to show that a metric space $(X,d)$ is totally bounded iff every sequence in $X$ has a Cauchy sub-sequence. This is a point that rose up the other day in another topic and it seemed like very nice thing to know.
The definition of total boundedness that I'm working with is that for every $\varepsilon >0$ there exists a finite collection $x_{1},...,x_{k}\in X$ so that $X=\cup_{n=1}^{k}B(x_{n},\varepsilon)$.
What I think I solved out so far is the proof from left to right, but I'm having trouble finishing the latter claim. Here's what I got from the first direction:
Suppose that $X$ is totally bounded and choose a sequence $(x_{n})_{n=1}^{\infty}$ of elements from $X$; we want to show that it has a Cauchy sub-sequence. Since $X$ is totally bounded there exists a collection $y_{1},...,y_{k_{1}}\in X$ such that $X=\cup_{n=1}^{k_{1}}B(y_{n},1)$, so there must exist an index $j_{1}\in\{1,...,k_{1}\}$ so that $x_{n}\in B(y_{j_{1}},1)=:A_{1}$ with infinitely many $n\in \mathbb{N}$. Total boundedness is hereditary so $A_{1}=\cup_{n=1}^{k_{2}}B(z_{n},\frac{1}{2})$ for some $z_{1},...,z_{k_{2}}\in A_{1}$. Again there exists $j_{2}\in \{1,...,k_{2}\}$ such that $x_{n}\in B(z_{j_{2}},\frac{1}{2})=:A_{2}$ for infinitely many $n\in \mathbb{N}$, and we will continue this process indefinitely. For each $i\in \mathbb{N}$ we find $A_{i}\subset X$ such that: $A_{i}=B(w_{i},\frac{1}{i})$ for some $w_{i}\in X$, $x_{n}\in A_{i}$ for infinitely many $n\in \mathbb{N}$ and $A_{i+1}\subset A_{i}$. For each $i\in \mathbb{N}$ we choose $x_{n_{i}}\in A_{i} \cap \{x_{n}\}_{n=1}^{\infty}$, which results a Cauchy sub-sequence $(x_{n_{i}})_{i=1}^{\infty}$. Adding few details and giving a more rigorous support for the last sentence, this should be fine?
I already tried several different things for the other direction, so I'd appreciate some fresh ideas there. Thanks for all the input in advance.