Let $X \subset R$
1) Compact => bounded.
I find it easy to just do this. For every $x \in X$ let $V_x = (x-1/2, x + 1/2)$. $V_x$ is open and $X \subset of \cup V_x$. So {$V_x$} is an open cover. So it has a finite subcover. So there is a lowest interval and there is a greatest interval in the finite subcollection of intervals and X is bounded between them.
2) Compact => closed
Let X not be closed. Then there is a limit point,y, of X that is not in X. Let's let $V_n$ = {$x \in \mathbb R| |x - y| > 1/n$}. As this covers all $\mathbb R$ except $y$ and $y \not \in X$ it covers X. Take any finite subcover the is a maximum value of $n$ so $(y - 1/n, y + 1/n)$ is not covered by the finite subcover. As $y$ was a limit point, $(y - 1/n, y + 1/n)$ contains points of X. So the subcover doesn't cover X. So X is not compact.
Unfortunately Closed and Bounded => compact is much harder.
But I hope I gave you a sense of the flavor of compact sets.
As I said in my comment, you are in good company---in fact, the company of Dedekind himself! In a letter to Heinrich Weber, Dedekind says the following:
(...) I would advise that by [natural] number one understand not the class itself (...) but something new (corresponding to this class) which the mind creates. (...) This is precisely the same question that you raise at the end of your letter in connection with my theory of irrationals, where you say that the irrational number is nothing other than the cut itself, while I prefer to create something new (different from the cut) that corresponds to the cut and of which I prefer to say it brings forth, creates the cut. (Ewald, From Kant to Hilbert, vol. 2, p. 835)
So Dedekind himself preferred not to identify the real number with the cut, merely saying that the mind somehow creates the real number which then corresponds to the cut. This is, however, a little bit obscure, so it's not surprising that most mathematicians (such as Weber!) decided to ignore Dedekind's suggestion and simply identify the real number with the cut. The reasoning behind this identification is roughly the following.
We know that any Dedekind-complete ordered field is isomorphic to the field of the real numbers. In particular, this means that any construction or theorem carried out in the real numbers could be reproduced inside an arbitrary Dedekind-ordered field, and vice-versa, by simply using the isomorphism as a "translation" between the fields. Hence, it doesn't matter what the real numbers actually are; for mathematical purposes, even supposing that there is such a thing as the real numbers, anything that we wanted to do with them could also be accomplished in an arbitrary Dedekind-complete ordered field.
Thus, if we could show that the cuts themselves satisfy the axioms for being a Dedekind-complete ordered field, then we could dispense with the real numbers altogether and simply work with the cuts themselves. And, in fact, we can show that this is the case! One need only to show that, given two cuts, $X$ and $Y$, it's possible to define operations on them corresponding to the usual operations on the real numbers, such as addition and multiplication, and that after doing so these operations will satisfy the field axioms. It's not difficult to see that the obvious operations will yield the desired result (exercise!), though it is somewhat laborious. If you are interested in seeing a detailed verification, I recommend reading, say, Appendix A of Yiannis Moshovakis excellent book Notes on Set Theory, which contains a very thorough discussion of the matter.
Best Answer
The first statement means that you might have a sequence of rational numbers which converge to a limit, but the limit is not rational. E.g., here's a sequence of rational numbers:
$3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ....$
in which each term is the first few digits of $\pi.$ The limit of this sequence is $\pi$, which is not rational.
The second statement says that the real numbers don't have this "flaw." Every convergent sequence of rational numbers (or for that matter, real numbers) converges to a real number. So the reals ARE closed under limits.