Prove $(X,d)$ is a metric space where $|X| < |\mathbb{R}|$, then $X$ is zero-dimensional.

Question

Can I please receive help with part (ii) of this problem? I have no clue how to solve it. It is from Munkree's topology book. Thank you! $\def\R{{\mathbb R}} \def\N{{\mathbb N}}$

A topological space $X$ is zero-dimensional if it has a basis $B$ consisting of open sets which are simultaneously closed.

For some reason, my solution is not posting, so I will try later to repost it.

Answer 1

For (i) use the form that is given: $C=\{0,1\}^{\Bbb N}$ which is a product of zero-dimensional spaces (a discrete space is of course zero-dimensional), and as such zero-dimensional too (use the standard base for a product where all non-trivial open sets come from a chosen clopen base in each factor space).

(ii) For any $x$, and any $r>0$, consider the function $f(y)=d(x,y): X \to \Bbb R$. $f[X]$ has size $< |\Bbb R|$ too, so cannot contain $(0,r)$ which has size continuum. So for some $0 < s < r$, the value $d(x,y)=s$ cannot occur for any $y$, which implies that $B(x,s)=\overline{B(x,s)} \subseteq B(x,r)$ and we have a clopen set inside every open ball of the same centre, which tells us $X$ is zero-dimensional.