I am attempting here to show the equivalence between the following three statements for the metric space $(X,d),$
i) $(X,d)$ is compact, meaning every open cover admits a finite subcover
ii) $(X,d)$ is sequentially compact, meaning every sequence in $X$ has a convergent subsequence
iii) $(X,d)$ is complete and totally bounded
Many of these proofs already exist on StackExchange, but the posts I could find were scattered and incomplete. This is an attempt to gather the proofs and to have someone $\underline{\text{check my reasoning and provide the remaining proofs I've been unable to do.}}$
If anyone knows shorter, more concise proofs, those would be appreciated! The post is already quite long as written.
Notation: $B_\rho(x)$ is an open ball of radius $\rho$ centered at $x,$ and $\mathbb{N}$ denotes the natural numbers.
i) $\implies$ ii): Let $\{p_i\}$ be an infinte sequence in $X.$ The set $S = \{p_1, p_2, \dots\}$ must have a cluster point, call it $p.$ If $S$ did not have a cluster point, then every open ball in $X$ would contain only a finite number of points. Since $X$ is compact and every cover has a finite subcover, this would imply that $S$ is not infinite. Since $S$ has a cluster point $p,$ pick $n_1 \in \mathbb{N}$ such that $p_{n_1} \in B_1(p).$ Then $n_2 > n_1$ such that $p_{n_2} \in B_{1/2}(p),$ and so on. Then $p_{n_k}$ for $k = 1,2,\dots$ will be a subsequence that converges because the n-th term will be less than $1/n$ away from $p.$
ii) $\implies$ i) : Let $C = \{O_1, O_2 , \dots\}$ be a countable open cover of $X,$ and suppose that there is no finite subcover that covers $X.$ Let $x_1 \notin O_1$ but $x_1 \in X,$ $x_2 \notin O_1 \cup O_2$ but $x_2 \in X$ and so on. One may always be able to pick such a point because no finite number of $O_i's$ cover $X.$ By hypothesis, the sequence $\{x_i\}_{i=1}^\infty$ must have a convergent subsequence that converges to $x \in X.$ Since $x \in X,$ $x \in O_n$ for some $n.$ But this would mean that an infinte number of points must be in that $O_n,$ which is not allowed by the construction of the sequence $\{x_n\},$ thus a contradiction.
i) $\implies$ iii) : Let $C$ be the open cover $\{B_\epsilon(x)| x \in X\},$ where $\epsilon > 0.$ Since $X$ is compact, a finite number of $B_\epsilon$'s cover $X,$ hence there exists an $\epsilon$ net for $X,$ meaning its totally bounded. Directly showing i) implies iii) seems excessive because i) implies ii) which trivially implies iii). I'd be interested to see a direct method if anyone knows one.
ii) $\implies$ iii) : Let $\epsilon > 0$ and assume that no finite collection of $B_\epsilon$ covers $X.$ Then for any finite collection of points in $X,$ namely $\{x_1, x_2, \dots, x_n\}$ there exists an $x \in X$ such that $d(x,x_i) > \epsilon$ for $i = 1, \dots, n.$ Let $y_n := x$ in this case. Then it is clear that the sequence $\{y_n\}$ has no convergent subsequence because $d(y_n,y_m) > \epsilon$ for all $n,m$ not equal. Completeness is trivially implied by ii), because if $X$ weren't complete, there would exist a Cauchy sequence in $X$ that didn't converge contradicting ii).
iii) $\implies$ ii) Letting $\{p_i\}$ be a sequence in $X,$ at least 1 $B_\epsilon$ contains an infinite number of points. One may then construct a nested sequence of $\epsilon$-balls, each of which contain an infinite number of $p_i$'s. Using this nested sequence to create a subsequence of $p_i$'s, completeness implies that this subsequence will in fact converge in $X.$
iii) $\implies$ i) Incomplete - I've had some trouble showing this result directly without first using $iii) \implies ii) \implies i).$ I'd be interested to see how its done.