The proof is correct, but it's rather non-minimal, since the extreme value theorem you use is stronger than the result you want and the proof for this specific case is easier than the proof of that theorem.
Assume the set didn't contain its supremum (which exists since the set is bounded). Since the set is closed, an entire neighbourhood of the supremum lies outside the set. This neighbourhood contains upper bounds of the set that are lower than the supremum, which is a contradiction.
It is important to understand that, although definitions often look arbitrary, they never are. Mathematical objects are intended to model something, and you can't understand why the definition is the way it is until you understand what it is trying to model. The question you asked is exactly the right one: why is it defined this way and not some other way? What is it trying to model?
(For example, why does a topology say that arbitrary unions of open sets are open, but infinite intersections of open sets might not be? It's because topology is intended to be an abstraction of certain properties of the line and the plane, and open sets are intended to be a more general version of open intervals of the line and open discs in plane, and that is how the intervals and discs behave.)
This case is similar. Mathematicians noticed that there are certain sorts of “well-behaved” subsets of the line and of metric spaces in general. For example:
- A continuous function is always uniformly continuous — if and only if its domain is well-behaved in this way
- A continuous real-valued function is always bounded — if and only if its domain is well-behaved in this way
- If $f$ is a continuous real-valued function on some domain, there may be some $m$ at which $f$ is maximized: $f(x) ≤ f(m)$ for all $x$. This is true of all such $f$ if and only if the domain is well-behaved in this way
- Every sequence of points from a subset of $\Bbb R^n$ contains a convergent subsequence — if and only if the subset is well-behaved in this way
and so on. It took mathematicians quite a long time to understand this properly, but the answer turned out to be that the "well-behaved" property is compactness. There are several equivalent formulations of it, including the open cover formulation you mentioned.
In contrast, the alternative property you propose, with closed covers, turns out not to model anything interesting, and actually to be trivial, as the comments point out. It ends nowhere. But even if it ended somewhere nontrivial, it would be a curiosity, of not much interest, unless it had started from a desire to better understand of something we already wanted to understand. It's quite easy to make up new mathematical properties at random, and to prove theorems about those properties, and sometimes it might seem like that is what we are doing. But we never are.
Properly formulated, compactness turns out to be surprisingly deep. Before compactness, mathematics already had an idea of what a finite set was. Finite sets are always discrete, but not all discrete sets are finite.
Compactness is the missing ingredient: a finite set is one that is both discrete and compact. With the discovery of compactness, we were able to understand finiteness as a conjunction of two properties that are more fundamental! Some of the properties we associate with finiteness actually come from discreteness; others come from compactness. (Some come from both.) Isn't that interesting?
And formulating compactness correctly helps us better understand the original space, $\Bbb R^n$ and metric spaces in general. Once we get compactness right, we see that the properties of "well-behaved" sets I mentioned above are not true of all compact spaces; metric spaces are special in several ways, which we didn't formerly appreciate.
Keep asking these questions. Every definition is made for a reason.
Best Answer
You've got part of the idea, but you need to pick your open sets to be distant from the supremum.
Let me propose, however, another argument. For each $x\in X$, the set $X_x=X\cap [x,\infty)$ is nonempty and closed, and if you intersect finitely many of these, you obtain
$$X_{x_1}\cap X_{x_2}\cap\dots\cap X_{x_n}=X_{\max \{x_i\}}\neq \emptyset.$$
Therefore by an equivalent formulation of compactness (which I strongly encourage you to review if you are not familiar with it, as it makes life easier sometimes to think of closed sets vs open), the intersection $$Y:=\bigcap_{x\in X} X_x\subseteq X$$ is nonempty. Let $y\in Y$, so that $y\in X$ as well. Then $y\in X_x$ for all $x\in X$ implies $y\geq x$ for all $x\in X$, so $y$ is the maximum element of $X$.
Remark
If we restate the problem to prove that a compact subset has a maximum, then the statement remains true in much greater generality than just $\mathbb R$, and in fact holds if we replace $\mathbb R$ with any totally ordered set with its order topology, using the exact same proof. (In this generality, closed and bounded need not imply compact, and in fact need not imply the existence of a maximum, so we must begin with compactness in the premise of the theorem.)