With the discrete metric $d(x,y) = 1$ for $x \ne y$, 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 $x \ne y$ is symmetric; and (M3) because $d(x,z)$ is $0$ or $1$, and when it is $1$ one has $x \ne z$, 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 $d(x_k, x_K) = 0 < \varepsilon$ for $k \ge K$ and every rational $\varepsilon > 0$, and $x_k \to x_K \in S$.

For claim 3, suppose $S$ is finite. If $S = \emptyset$ 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 $(1/2)$-net $F$; then $S = \bigcup_{y \in F} B(y,1/2) = \bigcup_{y \in F} \{y\} = F$ 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.