Proving total boundedness of a subset of a metric space This preface of this question is related to another question I posted, but what I am proving is quite different. So I am starting a new question. 
Let $\mathbb{R}^{\infty}$ be the infinite Cartesian product of $\mathbb{R}$ with itself. Define the metric $d(x,y) = \sum_{t \in \mathbb{N} \cup \{0\}} \beta^{-t} \rho(x_t, y_t)$ where $\rho(x_t, y_t) = \min\{|x_t - y_t|, 1\}$ for some fixed $\beta>1$ and for all $x, y \in \mathbb{R}^{\infty}$ where $x = (x_0, x_1, \cdots)$ and $y=(y_0, y_1, \cdots)$. Prove that $[0,1]^{\infty}$ is a totally bounded subset of $\mathbb{R}^{\infty}$ under $d$.
The definition of total boundness that I'm working with is: 
A set $A$ in a metric space $(X, d)$ is totally bounded if for every $\varepsilon>0$, there exists $x_1, \cdots, x_n \in A$ such that $A \subseteq \cup_{i=1}^{n} B_{\varepsilon}(x_i)$.
How can I use the above definition to prove that $[0,1]^{\infty}$ is totally bounded? So what I need to do is find $x_1, x_2, \cdots, x_n \in [0,1]^{\infty}$ where $x_1 = (x_{1, 0}, x_{1, 1}, \cdots)$, $x_2 = (x_{2, 0}, x_{2, 1}, \cdots)$, so that the finite union of the open balls around these elements contains $[0,1]^{\infty}$. But I have no idea how to find these particular $x_1, x_2, \cdots, x_n$ to complete the proof. Any help would be appreciated.
 A: Let's gather up a couple of things that are useful:


*

*For $x,y\in[0,1]^\infty$, we know that $\rho(x_t,y_t)\leq 1$.

*$$\sum_{t=0}^\infty \beta^{-t}\rho(x_t,y_t)\leq \sum_{t=0}^\infty \beta^{-t} = \frac{1}{1-\beta}=:B$$

*For every $\epsilon>0$, there exists $T$ such that $$\sum_{t=T}^\infty \beta^{-t} < \epsilon$$



Now, let $\epsilon > 0$, let $T$ be such that $$\sum_{t=T}^\infty \beta^{-t} < \frac\epsilon2$$ and take $x^1=(0,0,0\dots)$.
Now, take any $x = (x_1,x_2,x_3,\dots)$ such that for $i<T$, you have $|x_i|<\frac\epsilon{2B}$. Then,
$$d(x,x_1)=\sum_{t=0}^\infty \beta^{-t} \rho(x^1_t, x_t) = \sum_{t=0}^\infty \beta^{-t}|x_i|\\=\sum_{i=0}^{T-1}\beta^{-t}|x_t| + \sum_{i=T}^\infty \beta^{-t}|x_t|$$
Now, for the first sum, use $|x_t|<\frac\epsilon{2B}$, and for the second, use the fact that $|x_t|\leq 1$ to get
$$d(x,x_1)\leq \frac{\epsilon}{2B}\sum_{i=0}^{T-1}\beta^{-t} + \sum_{t=T}^\infty \beta^{-t}\\
\leq \frac{\epsilon}{2B}\sum_{i=0}^\infty \beta^{-t} +\sum_{t=T}^\infty \beta^{-t}\\
\leq \frac\epsilon{2B}\cdot B + \frac{\epsilon}{2} = \epsilon$$
So, we have shown that $x_1\in B(x^1,\epsilon)$

The take-away:
$x$ is only bounded on the first $T$ coordinates, and it can be anything for the other coordinates. Therefore, in a way, you only need to split $\mathbb R^T$ in a sort of chessboard-style pattern to get the full coverage. For example, you would take $x^2=(\frac{\epsilon}{2B}, 0,0,0\dots)$, and in general (taking $\delta=\frac{\epsilon}{2B}$) you would take $x_n$ to form the set
$$\left\{\left(k_1\delta, k_2\delta,k_3\delta,\dots, k_T\delta, 0,0,0\dots\right)|k_1,k_2\dots, k_T\in\left\{0,1,2,\dots\left\lceil\frac{1}{\delta}\right\rceil\right\}\right\}$$
which is clearly finite. The proof that this is a good selection is similar to the proof of how $x^1$ covers the set 
$$\{(x_1,x_2\dots| x_1\leq\delta, x_2\leq\delta,\dots, x_T\leq\delta\}$$
