Please help me understand this proof of exercise in Rudin PMA:
Exercise Let $f$ be real uniformly continuous function on the bounded set $E$ of $\Bbb R$. Prove that $f$ is bounded on $E$.
Proof Let $a=\inf E$ and $b=\sup E$, and let $\delta >0$ be such that $|f(x)-f(y)|<1$ if $x,y\in E$ and $|x-y|<\delta$. Now choose a positive integer $N$ larger than $(b-a)/\delta$, and consider the $N$ intervals $I_k=\left[a+\dfrac{k-1}{b-a},a+\dfrac{k}{b-a}\right], \,k=1,2,\ldots,N.$ For each $k$ such that $I_k\cap E=\varnothing$ let $x_k\in E\cap I_k$. Then $M=1+\max\{|f(x_k)|\}$. If $x\in E$ we have $|x-x_k|<\delta$ for some $k$, hence $|f(x)|<M$.
I can't understand why $N$ chose so that $N>(b-a)/\delta$.
Also, don't know how $x\in E$ implies $|x-x_k|<\delta$ for some $k$.
It's to be sure to have that $|I_k|<\delta$, and thus, if $x,y\in I_k$, then $$|x-y|<\delta\implies |f(x)-f(y)|<1.$$