TITLE: What does it REALLY mean for a metric space to be compact?
QUESTION [13 upvotes]: I've been trying to wrap my head around the concept of compactness and get an intuitive understand of what it is. The definition used in my text book is the finite subcover definition.

A subset $K$ of a metric space $X$ is said to be compact if every open cover of $K$ contains a finite subcover. More explicitly, the requirement is that if {$G_{\alpha}$} is an open cover of $K$, then there are finitely many indices $\alpha_1,...,\alpha_n$ such that $$K\subset G_{\alpha_{1}}\cup\, ...\cup \, G_{\alpha_{n}}.$$

This definition is not very accessible to me so I've been looking around trying to find something to help me understand it. 
So far, I haven't really wrapped my head around the idea yet, but I have learned the following:

Compactness is a kind of limited-ness.
Compactness is one of the two properties of finiteness, the other being discreteness. (I saw this in the explanation about foos, the creatures that are red and short, and the word foo has come to mean something both red and short.) 
In $R^k$, compactness is equivalent to being closed and bound. 

So I guess my question is, what is it about compactness that led mathematicians to call it "compact"? What exactly is compact about it? What does this have to do with the definition (that is, where does the definition come from)? Furthermore, what does it mean to be discrete? I think it would help if you could give me an example of metric spaces that are:

Compact and discrete
Compact but not discrete
Discrete but not compact
Neither discrete or compact

I've already read this question: What should be the intuition when working with compactness?
The answers on this post explain very well why it is difficult to understand compactness, but I was hoping for something more concrete to help me understand.

REPLY [2 votes]: I will try my best to motivate that definition.
Let's say $y$ is "close" to $x$ if the distance between $x$ and $y$ is less than $1$. Now, we know that — with this definition of "close" — everything in $\Bbb R$ is "close" to something in $\Bbb Z=\{\dots,-2,-1,0,1,2,\dots\}$. Also, it's quite easy to see that we can't replace $\Bbb Z$ with a finite set. With this definition of "close," $\Bbb R$ is not "close" to any finite set.
What about $(0,1)$? Well, with this definition of "close," everything is close to $\frac12$. However, what if we change our definition of close? We can make the definition of "close to $x$" depend on $x$. For example, let's say that $y$ is "close" to $x$ if $y\in(\frac x2,1)$. (This isn't a symmetric relation anymore, sorry.) Now, can we say that everything in $(0,1)$ is "close" to some finite set? It's not hard to show that, now, no finite set works. We can see that everything in $(0,1)$ is close to something in $\{\frac12,\frac14,\frac18,\dots\}$, but we can't say it's "close" to some finite set.
So there's more than one possible definition of "close." Can anything be a definition of "close"? No, that would be too boring. One thing we know should be true about the definition of "close": the set of things "close" to $x$ should be some open set containing $x$. (I don't want the set of points "close" to $0$ to be $[0,1)$, for example, since this means that not a single negative number is "close" to $0$.) I don't care about anything else; as long as it satisfies this, it's a plausible definition of "close" to me.
And $[0,1]$? This is harder to prove, but it turns out that no matter what definition of "close" you think of, I can find a finite set $A$ such that everything in $[0,1]$ is "close" to it. Which leads us to:

A space $X$ is called compact if, for every definition of "close" you can think of (in the sense above), there is a finite set $A$ such that everything in the space is close to something in $A$.

This ends us being a useful definition, because now, instead of studying the whole (potentially infinite) space $X$, we usually only really need to pay attention to a finite subset. However, in this formulation, it's not too useful. It doesn't look formal enough, honestly.
How can we make this more formal? Well, let a definition of "close" be agreed on, and let $O_x$ be the set of points "close" to $x$. Clearly, $\{O_x:x\in X\}$ covers $X$ (since every point $a$ is covered by $O_a$). If the space was compact, the definition above says there'd be a finite set $A$ such that $\{O_x:x\in A\}$ covered $X$. That is, there would be a finite subcover. If the space wasn't compact, there'd be a definition of "close" that doesn't work — and hence a cover $\{O_x:x\in X\}$ with no finite subcover.
All this leads to our equivalent definition:

A space $X$ is compact if every open cover has a finite subcover.