I'm having trouble writing down the details of this proof formally.

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> **Statement**: Suppose $(X, d)$ is a metric space that is *complete*, and *totally bounded* (i.e., for every $\epsilon > 0$, $\exists$ finitely many points $x\_{i} \in X$, $i = 1, \dots, n$ such that $X = B(x\_{1}, \epsilon) \cup \dots \cup B(x\_{n},\epsilon) )$. Prove that $X$ is *sequentially compact*.
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Here is the idea of the proof:

Let $\{ y\_{n} \}\_{n=1}^{\infty}$ be a sequence in $X$. We want to show we can find a convergent subsequence. But the space is totally bounded, so for $\epsilon = \frac{1}{2}$, we can find finitely many $x\_{i} \in X$ ($i = 1, \dots, n $) such that $X = B(x\_{1}, \frac{1}{2}) \cup \dots \cup B(x\_{n},\frac{1}{2}) $. Then infinitely many terms of $\{y\_{n} \}$ appear in some $B(x\_{i}, \frac{1}{2})$.

If we look at the subsequence of infinitely many terms from $\{y\_{n} \}$ that appear in $B(x\_{i}, 1)$, it's clear the distance between each two terms in this subsequence is less than $1$. I'll call this subsequence $\{y\_{n}^{(1)} \}$.

We can repeat this process with $\epsilon = \frac{1}{2^{2}}$, that is, find finitely many open balls the union of which equals $X$, and one of these balls contains infinitely many terms of $\{ y\_{n}^{(1)} \}$. I'll call this subsequence $\{ y\_{n}^{(2)} \}$, and clearly the distance between every two terms in this sequence is less than $\frac{1}{2}$.

Doing this iteratively, I can then pick an element from each of these subsequences I am finding, and this will construct my Cauchy subsequence of the original sequence, which will converge by the completeness of the space.

It's clear that from this subsequence, the distance between two terms, say, $y\_{n}^{(m)}$ and $y\_{n}^{(p)}$ is less than $\frac{1}{2^{p}}$ if $p < m$.

So for each $\epsilon > 0$, all I need to do is find $N$ such that $\frac{1}{2^{N-1}} < \epsilon$. Then $t, s \geq N$ implies $d(y\_{n}^{(t)}, y\_{n}^{(s)})< d(y\_{n}^{(t)}, y\_{n}^{(N)})+ d(y\_{n}^{(N)}, y\_{n}^{(s)}) < \frac{1}{2^{N}} + \frac{1}{2^{N}} < \frac{1}{2^{N - 1}} < \epsilon $.

Did I manage to keep all of the details straight? Thanks for any help.