Metric space which is totally bounded is separable. Baby Rudin Ex 2.24

Question

Let $X$ be a metric space which is totally bounded. Show that $X$ is separable.

A metric space is called separable if it contains a countable dense subset.

A metric space is called totally bounded if for every $\delta>0$ there exists $N_{\delta}\in \mathbb{N}$ and $x_i\in X, \, 1\le i\le N_{\delta}$ such that $$X \subseteq \bigcup_{i=1}^{N_\delta} B(x_i,\delta).$$

My attempt :

Since $X$ is totally bounded for each $\delta_n = 1/n$ there exists a finite subset $A_n$ of $X$ containing $N_{\delta_n}(\in \mathbb{N})$ elements such that $$X \subseteq \bigcup_{x\in A_n} B(x,1/n).$$

Let $$A = \bigcup_{i=1}^{\infty} A_i.$$

Claim : A is a countable dense subset of $X.$

$A$ is dense :

Consider $B(x,\epsilon)$ for any $x\in X$ where $\epsilon>0.$ Since there exists some $p\in A_k$ such that $x\in B(p,1/k)$ and $k>1/\epsilon$ implies that $x$ is either a limit point of $A$ or a point of $A.$ Hence $A$ is dense

$A$ is countable :

Now this is where my trouble begins. Since Rudin defines that a set is countable if it is in one-one correspondence with the positive integers. Also he defines a set to be at most countable if it is finite or countable.

Here since $A$ is a union of finite sets, $A$ is at most countable.

Hence to show that $A$ is countable I need only show that $A$ cannot be finite.

So this how I proceeded . . .

Suppose $A$ is finite. This would mean (or would this ?)$$A = A_{m_1}\cup A_{m_2}\cup \cdots \cup A_{m_k} (?)$$

Now I think maybe by considering some large $m$ and invoking Archimedean property, I might get the contradiction I'm looking for.

Can anyone please help me complete this?

Answer 1

As I noted in the comments:

Most authors use the term countable to mean finite or countably infinite. As you indicate, Rudin uses the term countable to mean countably infinite. Based on that, it appears the exercise accidentally slips back to the more common usage (else a finite metric space would be a counterexample).

But to answer your followup question . . .

Yes, if $X$ is infinite, then the set $A$ will also be infinite.

Suppose instead that $A$ is finite.

Choose $y\in X{\,\setminus\,}A$.

Since $A$ is dense, there is an infinite sequence of elements of $A$ which approach $y$.

But then since $A$ is finite, some element of the sequence has minimum distance to $y$, contradicting convergence.