How to prove a set is a totally bounded set?

Question

Let $(S,d)$ be a separable complete metric space. By separability, for every $n\ge 1$ there exists an open $1/n$ ball sequence $A_{nk}=\{x\in S: d(x,x_k)<1/n\}$ covering $S$, where $x_k\in S, k\ge 1$. $k_n$ is a positive integer depending on $n$. How to prove that $B=\bigcap_{n\ge 1}\bigcup_{k\le k_n}A_{nk}$ is a totally bounded set.

Answer 1

I obtain the following proof, please correct it for me. Thank you. Proof. For any $\varepsilon>0$ there exists a positive integer $n'$ such that $1/n'<\varepsilon/2$. Writing $A_{n'_k}=B(z_k, 1/n')$, and noting that $B\subset\bigcup_{k\le k_{n'}}A_{n'_k}$, we have $B=\bigcup_{k\le k_{n'}}\left(B\bigcap A_{n'_k}\right)$. Hence, for every $x\in B$, there exists $k_0\in \{1,2,\cdots,k_{n'}\}, x\in A_{n'_{k_0}}$. Taking $y_{k_0}\in B\bigcap A_{n'_{k_0}}$, then $$d\left(x, y_{k_0}\right)\le d\left(x, z_{k_0}\right)+d\left(y_{k_0}, z_{k_0}\right)<1/n'+1/n'<\varepsilon.$$ So, $\tilde{B}=\{y_{k_0}, 1\le k_0\le k_{n'}\}$ is a finite $\varepsilon$-net for $B$. Thus $B$ is a totally bouned set. This completes the proof.