= Solution
A <metric space> $X$ is a <compact space> if every <open cover> $X=\bigcup_{\alpha\in I}U_\alpha$, with all $U_\alpha$ open in $X$, has a <finite subcover>: finitely many indices $\alpha_1,\ldots,\alpha_r$ cover $X$.
It is a <sequentially compact space> if every <sequence> $(x_n)$ in $X$ has a <convergent subsequence> $(x_{n_k})$ whose <limit of a sequence> belongs to $X$. In formulas,
$$
\boxed{\forall(x_n)\subseteq X\ \exists x\in X,\ n_1<n_2<\cdots:\ d(x_{n_k},x)\to0}.
$$
A <finite subcover> is the defining test for <compactness>; a <convergent subsequence> is the defining test for <sequential compactness>.
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