Understanding compact subsets of metric spaces Please help me understand the following definition:

Let $(X,d)$ be a metric space, a subset $S \in X$ is called compact, if any infinite sequence $\{x_{n}\}_{n\in\Bbb N}\in S$ has a sub-sequence with a limit in S.



*

*What does "if any infinite sequence" mean? Maybe: At least one, all?

*What does "has a sub-sequence" mean? Maybe: At least one? Exactly one?
I am no mathematician and I don't understand the (practical) relevance of this property. Please explain it to me.
 A: The definition can be reformulated as follows - "Let $(X,d)$ be a metric space, a subset $S∈X$ is called compact, if for all infinte sequences $\{ x_{n}\}_{n=1}^{∞}\subseteq S$ the following holds: $\{ x_{n}\}_{n=1}^{∞}$ has a concentration point and if $\bar{x}$ is a concentration point of $\{x_{n}\}_{n=1}^{∞}$ , than $\bar{x} \in S$." Now consider the definition of a concentration point of a infinite sequence. The practical side of compactness of a given set is that it contains it's "edge". If any definition seem vague to you, try to rewrite it using relevant notions with which you're more familiar with.
A: 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:


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*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.
A: Compactness is a subtle "finiteness" property. Unfortunately there is no simple way to characterize resp. to define it in general. Only for subsets $S\subset {\mathbb R}^n$ there is a simple characterization: $S$ has to be closed and bounded. Bounded means, of course, that $S$ should fit in a  ball of finite radius, while closed means that the boundary $\partial S$ is included in $S$.
Now the practical consequences of being compact are huge. The simplest consequence is that a a continuous function $f:\ S\to{\mathbb R}$ on a compact set $S$ is automatically bounded, i.e., there is an $M$ such that $|f(x)|\leq M$ for all $x\in S$. But more is true: You are guaranteed at least one point $\xi\in S$ where $f$ takes its global maximum: There is a $\xi\in S$ with $f(\xi)=\max_{x\in S}f(x)$. 
