I have trouble understanding the following proof:
Consider the metric space $([-1,1]^d, \|\cdot\|_\infty)$. Let $\epsilon >0$. Fix $n\in\mathbb{N}$ such that $\dfrac{1}{n} <\epsilon$. Let $I_n = \{-n, -n+1, \ldots, n-1, n\}$. Now $B_\epsilon\bigg(\dfrac{k_1}{n},\ldots,\dfrac{k_d}{n}\bigg):k_1,\ldots,k_d\in I_n$ is a finite collection of balls of radius $\epsilon$ which has a union which contains $[-1,1]^d$. Indeed, if $x\in[-1,1]^d$, then we can find $k_1,\ldots,k_d\in I_n$ such that $k_j\leq x_j\leq k_{j+1}$. Now one can check that $x\in B_\epsilon\bigg(\dfrac{k_1}{n},\ldots,\dfrac{k_d}{n}\bigg)$ and we have that $([-1,1]^d,\|\cdot\|_\infty)$ is totally bounded.
Question: How would you show that $B_\epsilon\bigg(\dfrac{k_1}{n},\ldots,\dfrac{k_d}{n}\bigg)$ has a union which contains $[-1,1]^d$? I find it really hard to imagine how many balls you would need to cover a set that can no longer be visualised. I've tried think about it in the following way:
Suppose that we just consider $(a,b)$, an interval on the real line $\mathbb{R}$. Given $\epsilon >0$, I think we would need $\dfrac{b-a}{\epsilon}$ balls with radius $\epsilon$ to cover the entire interval. Consider now $(a,b)\times (c,d)$ on $\mathbb{R}^2$. Given $\epsilon$, I think we would need $\dfrac{b-a}{\epsilon}\times \dfrac{d-c}{\epsilon}$ balls with radius $\epsilon$ to cover $(a,b)\times (c,d)$. If I now take a look at $[-1,1]^d$, this reasoning would mean that it would take $\bigg(\dfrac{2}{\epsilon}\bigg)^d$ balls with radius $\epsilon$ to cover $[-1,1]^d$. What would the union of these balls look like?