The proof one usually sees for this (by my knowledge) is by proving that the metric space in question is sequentially compact, as seen here:
totally bounded, complete $\implies$ compact
I am interested in a more direct proof involving the open covering definition of compactness. Using the equivalence of countable compactness and limit point compactness (every infinite subset admits a limit point) for $T_1$-spaces as a starting point, I came up with the following proof, but it still feels a bit indirect.
Is anyone aware of a different and/or more direct proof? Thank you!
Here is my proof:
Assume $X$ to be a complete and totally bounded metric space. To prove that it is compact, we need to show that it is both countably compact and Lindelöf and both of these properties are equivalent to limit point compactness and separability, respectively, so we prove those two properties.
To see that $X$ is separable, note that for each $n\in\mathbb{N}$, there exists a finite $(1/n)$-net $\{x_{n,1},\dotsc,x_{n,r_n}\}$ for $X$. Then the countable set
\begin{equation}
S=\bigcup_{n=1}^{\infty}\{x_{n,1},\dotsc,x_{n,r_n}\}
\end{equation}
is dense in $X$, which proves that $X$ is separable.
To prove that $X$ is limit point compact, let $F\subset X$ be an infinite subset. Then as $X$ is covered by finitely many balls of radius $1$, it follows that one such ball contains infinitely many members of $F$, i.e. $F\cap B_{1}(x_1)$ is infinite for some $x_1\in X$. Similarly, as $F\cap B_{1}(x_1)$ is infinite and $X$ is covered by finitely many balls of radius $1/2$, we have that $F\cap B_{1}(x_1)\cap B_{1/2}(x_2)$ is infinite for some $x_2\in X$. Continuing inductively, we obtain a sequence $\{x_n\}_{n=1}^{\infty}$ in $X$ such that
\begin{equation}
E_n:=F\cap B_{1}(x_1)\cap\dotsb\cap B_{1/n}(x_n)
\end{equation}
is infinite for every $n\in\mathbb{N}$. I claim that $\{x_n\}_{n=1}^{\infty}$ is Cauchy. If $y_n\in E_n$ is arbitrary for $n\in\mathbb{N}$ then $\{y_n\}_{n=1}^{\infty}$ is Cauchy since the $E_n$ are nested and $\mathrm{diam}(E_n)\leq 2/n$, which in turn implies that $\{x_n\}_{n=1}^{\infty}$ is Cauchy as $d(x_n,y_n)<1/n$ for all $n$. Thus $\lim_{n\to\infty}x_n=x$ for some $x\in X$ by completeness. It remains to show that $x$ is a limit point of $F$, for which it is sufficient to show that $F\cap B_{\varepsilon}(x)$ is infinite for all $\varepsilon>0$.
To see this, let $\varepsilon>0$. Choose $n\in\mathbb{N}$ such that $1/n<\varepsilon/2$ and $d(x_n,x)<\varepsilon/2$. Then $B_{1/n}(x_n)\subset B_{\varepsilon/2}(x_n)\subset B_{\varepsilon}(x)$, so we obtain
\begin{equation}
E_n\subset F\cap B_{1/n}(x_n)\subset F\cap B_{\varepsilon}(x).
\end{equation}
Hence $F\cap B_{\varepsilon}(x)$ is infinite as $E_n$ is, which completes the proof.
Best Answer
You can prove this theorem directly using the open cover definition of compactness, but it is a proof by contradiction. I am sketching the main points here: