First, you’re right that a totally bounded metric space is automatically separable, so that there’s no need to go through the completion: the union of finite $2^{-n}$-nets for $n\in\omega$ is a countable dense subset.
Now assume that $\langle X,d\rangle$ is a separable metric space. Without loss of generality assume that $d(x,y)\le 1$ for all $x,y\in X$, and let $D=\{x_n:n\in\omega\}$ be a dense subset of $X$. Define the map
$$f:X\to[0,1]^\omega:x\mapsto\big\langle d(x,x_n):n\in\omega\big\rangle\;.$$
Now show that $f$ is an embedding of $X$ into the compact metrizable space $[0,1]^\omega$, the Hilbert cube; being compact, the Hilbert cube is totally bounded in any compatible metric, and total boundedness is hereditary, so $f[X]$ is totally bounded in any metric inherited from $[0,1]^\omega$.
For each $n \in \mathbb{Z}_{> 0}$, we take balls $B^n_1, \ldots, B^n_{n_m}$ of radius $1/n$ with centers $x^n_1, \ldots, x^n_{m_n}$ as you have described. Call the collection of centers $\mathcal{C}_n = \{x^n_1, \ldots, x^n_{m_n}\}$. Then $\mathcal{C} = \bigcup_{n \geq 1} \mathcal{C}_n$ is a countable union of finite sets, hence is itself countable. We claim this is a countable dense subset. To show this, your proof should start with: "Given $\epsilon > 0$, choose $n$ such that $1/n < \epsilon$, and consider the centers of the balls $x^n_1, \ldots, x^n_{m_n} \in \mathcal{C}_n$." Do you see how this implies that $\mathcal{C}$ is dense? Note that we defined $\mathcal{C}$ before we ever mentioned $\epsilon$.
Best Answer
In the second link the inverse map is continuous: If $d(x^{j},x_n) \to d(x,x_n)$ for every $n$ then then $d(x^{j},x)\to 0$.
Indeed, $d(x^{j},x)\leq d(x^{j},x_n)+d(x_n,x)$. First choose $n$ such that the second term is less than $\epsilon /2$ and then choose $m$ such that the first term is less than $\epsilon /2$ for $j \geq m$.