Every closed interval in $\mathbb{R}$ is totally bounded

Question

Let ($\mathbb{R}$,|.|) be the real metric space. Let $a\lt b \in \mathbb{R}$. Show that $([a,b],|.|)$ is totally bounded ( knowing that (X,d) is totally bounded if $\forall \varepsilon \gt0 $ , $\exists x_{1},...,x_{n}\in X$ such that $ X\subseteq \bigcup_\limits{i\in {1,...,n}} B(x_{i},\varepsilon) $ ). Let $\varepsilon \gt0$ there exists $n\in \mathbb{N}$ such that $n\varepsilon \gt (b-a) $ Now $\forall k\in {0,...,n} $ Let $x_{k}=a+k\frac{(b-a)}n$. Let's show that $[a,b]\subseteq \bigcup_\limits{k\in {0,...,n}} B(x_{k},\varepsilon)$. Let $x\in[a,b]$. If $x=a$ we have $x\in B(x_{0},\varepsilon)$ And if $x=b$ we have $x\in B(x_{n},\varepsilon)$ .. I'm stuck at the point where $x\in]a,b[$ .. Any help please.

Answer 1

if $x\in (a,b)$ take the smallest $x_i$ such that $x\leq x_i$. It follows $x_i-\frac{b-a}{n}<x\leq x_i$ and so $|x_i-x| = x_i-x < \frac{b-a}{n} < \epsilon$.