Recently in our UCLA summer analysis course we discussed the equivalence of compactness and sequential compactness for subsets of a metric space. I thought of a proof which I thought to be considerably shorter and simpler than those usually given. However it requires Zorn's Lemma, so it is perhaps not as fundamental.
Theorem. A sequentially compact subset of a metric space is compact.
Proof. Suppose $A \subset M$ (where $M$ is a metric space) has an open cover with no finite subcover. Let this cover be given by $C=\{O_i\}_{i\in I}$, where $I$ is an arbitrary indexing set (could be uncountable). Subcovers of $C$ are partially ordered by inclusion, and every chain of subcovers has a minimal element (given by the intersection of all of its elements). Thus by Zorn's Lemma we can assume that ${C}$ has a minimal subcover, which we will call ${C}'=\{O_k\}_{k\in K}$.
We now argue that each $O_k \in C'$ has an element $x_k$ such that $x_k \in {O}_j$ if and only if $j=k$. Indeed if we had an open set $O_k$ such that every element $x \in O_k$ was contained in some other open set in $C'$, the collection ${C}''={C}'-\{O_k\}$ would be an open cover and obviously a proper subset of ${C}'$, contradicting the assumption that ${C}'$ is a minimal subcover.
By the axiom of choice ${C}'$ has a countable subset $\{O_1,\ldots \}$. Out of each $O_i$ we pick an element $o_i$ such that $o_i \in O_j$ if and only if $i=j$ (guaranteed to exist by previous paragraph). Thus $\{o_i\}_{i=1}^\infty$ is a sequence. It could not have a convergent subsequence because then the limit point $o$ would be in some set $O_m \in {C}'$, which would then necessarily contain some $o_i$ with $i \ne m$.
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