With the discrete metric $d(x,y)=1$ for $x\mathrm{\ne }y$, a space is compact iff it is totally bounded iff it is finite, and it is complete whatever its size

$d$ is a metric: (M1) holds by definition; (M2) because the condition $x\mathrm{\ne }y$ is symmetric; and (M3) because $d(x,z)$ is $0$ or $1$, and when it is $1$ one has $x\mathrm{\ne }z$, so $y$ differs from at least one of $x$ and $z$ and $d(x,y)+d(y,z)\ge 1$.

For $0<r\le 1$, $d(x,y)<r\le 1$ forces $d(x,y)=0$ and $y=x$, so $B(x,r)=\{x\}$; for $r>1$ every $y$ has $d(x,y)\le 1<r$, so $B(x,r)=S$. Hence every $A\subseteq S$ is open, each $x\in A$ having $B(x,1)=\{x\}\subseteq A$, and every subset is closed as well, its complement being open: claim 1.

Claim 2: let $(x_k)$ be Cauchy and take $K$ with $d(x_m,x_n)<1$ for $m,n\ge K$; then $d(x_m,x_K)=0$, that is $x_m=x_K$, for every $m\ge K$, so $d(x_k,x_K)=0<\epsilon$ for $k\ge K$ and every rational $\epsilon >0$, and $x_k\to x_K\in S$.

For claim 3, suppose $S$ is finite. If $S=\mathrm{\varnothing }$ then the empty subfamily of any family covers it; otherwise list $S=\{s_0,\dots ,s_p\}$, let $\mathcal{U}$ be a family of open sets with union $S$, and note that for each $i\le p$ the set $\{U\in \mathcal{U}:s_i\in U\}$ is nonempty, so finite choice supplies $U_0,\dots ,U_p\in \mathcal{U}$ with $s_i\in U_i$; their union contains every $s_i$ and hence is $S$. So $(S,d)$ is compact.

If $(S,d)$ is compact it is totally bounded.

If $(S,d)$ is totally bounded, take a finite $(1\mathrm{/}2)$-net $F$; then $S={\bigcup }_{y\in F}B(y,1\mathrm{/}2)={\bigcup }_{y\in F}\{y\}=F$ by step 2.1, so $S$ is finite.

Steps 3.1, 4.1 and 5.1 close the cycle finite $\Rightarrow$ compact $\Rightarrow$ totally bounded $\Rightarrow$ finite, so the three conditions of claim 3 are equivalent.