The definition that you quoted is in fact that of sequential compactness. A subset $K$ of a topological space $X$ is called compact (see I J Maddox, p.62) if any open cover has a finite sub cover. Precisely, if $\{G_{\alpha}\}$ is a collection of open sets that covers $K$, then there exists a finite collection $G_{\alpha_1},G_{\alpha_2},...,G_{\alpha_n}$ which covers $K$.
Now a $metric$ space is said to be sequentially compact if and only if every sequence has a convergent subsequence. There is a theorem that establishes a link between the two: a metric space is compact if and only if it is sequentially compact (Maddox, Theorerm 21).
Towards your points of concern:
- Any sequence means any sequence, It is not sufficient to prove this for one particular sequence.
2."Has a sub-sequence" means "at least one". Once you have found one, you can extract a convergent sub-sub-sequence from it.
Compact spaces are useful, because they allow one to construct convergent sequences which are themselves of paramount importance in proving many results.
For example, there is a theorem in approximation theory (see M J D Powell, Theorem 1.1.) that says that asserts the existence of the best approximation to an element of a linear space from a compact subspace.