How to understand compactness? [duplicate]

Maybe you should think about compactness, as something that takes local properties to global properties. For example, if $f:K\rightarrow \mathbb{R}$ is continuous, $K$ is compact, and $f(x)>t_x>0$ for all x, then you can find $t>0$ such that $f(x)>t>0$ for all x - so from $f(x)>t_x>0$ point wise, you know that $f>t>0$ as a function. (This is a simple consequence of Weierstrass theorem in $[a,b]\rightarrow \mathbb{R}$)

Usually we find some property that is true for every "small" enough open sets, then use compactness to reduce the case to finitely many open sets and use induction to show that the property is true for all of the space. This is at least how I understand compactness.

As the commenters below this message wrote (and I didn't emphasize enough), we usually use compactness to reduce infinite problems\conditions\restraints to a finite subset that cover the entire space, and then use some argument that works only for finite cases (like induction, taking max\min, take finite sums etc). In my example above, we wanted to find a minimum over all the lower bounds, but in the infinite case this is usually just an infimum (and can be zero), but when we reduce to a finite case there is a minimum.

In this way we can think of compactness as something that let us use some finite argument on infinite covers (and many times to transfer some property from the cover to the entire space).

As I mentioned in another question, I tend to intuit compactness as a sort of "super closedness." There are a few reasons for this.

First, in Hausdorff spaces, every compact subset will automatically be closed.

Second, in first-countable spaces, one can think about "closedness" as "closed under limit operations." That is, a set $E$ is closed if and only if it contains the limits of all sequences in $E$. By analogy, every compact subset of a first-countable space is sequentially compact -- that is, every sequence has a convergent subsequence. (Note, however, that the converse of this is not generally true in first-countable spaces.)

Third, in metric spaces, compactness is equivalent to being complete and totally bounded. This can be viewed as a generalization of the Heine-Borel Theorem which classifies the compact sets of $\mathbb{R}^n$. In such a scheme, completeness corresponds to closedness, and total boundedness to boundedness.

Note that metric spaces are automatically Hausdorff and first-countable, so all of the above certainly applies to metric spaces.

For general topological spaces (without any assumption of "Hausdorffness" or first countability), one can also intuit compact topological spaces as those with relatively few open sets. Indeed, if a set $X$ is compact under a given topology, then it will also be compact under any weaker (coarser) topology.

EDIT: To add to Prometheus' answer, I would also say that compactness is a property that plays especially well with continuity (Extreme-Value Theorem, Heine-Cantor Theorem) and convergence of functions (Dini's Theorem). As Prometheus said, this is in part because of its role in reducing global problems to local ones -- or, perhaps, infinite ones to finite ones.

(I'm thinking particularly of the proof of Heine-Cantor which relies on the fact that the infimum of a finite number of positive distances is positive -- whereas an infinite number of positive distances may have an infimum of zero.)