I think that what you’re missing is that an open cover of a compact set can cover more than just that set. Let $X$ be a topological space, and let $K$ be a compact subset of $X$. A family $\mathscr{U}$ of open subsets of $X$ is an open cover of $K$ if $K\subseteq\bigcup\mathscr{U}$; it’s not required that $K=\bigcup\mathscr{U}$. You’re right that $\bigcup\mathscr{U}$, being a union of open sets, must be open in $X$, but it needn’t be equal to $K$.
For example, suppose that $X=\Bbb R$ and $K=[0,3]$; the family $\{(-1,2),(1,4)\}$ is an open cover of $[0,3]$: it’s a family of open sets, and $[0,3]\subseteq(-1,2)\cup(1,4)=(-1,4)$. And yes, $(-1,4)$ is certainly open in $\Bbb R$, but $[0,3]$ is not.
Note, by the way, that it’s not actually true that a compact subset of an arbitrary topological space is closed. For example, let $\tau$ be the cofinite topology on $\Bbb Z$: the open sets are $\varnothing$ and the sets whose complements in $\Bbb Z$ are finite. It’s a straightforward exercise to show that every subset of $\Bbb Z$ is compact in this topology, but the only closed sets are the finite ones and $\Bbb Z$ itself. Thus, for example, $\Bbb Z^+$ is a compact subset that isn’t closed.
It is true, however, that compact sets in Hausdorff spaces are closed, though a bit of work is required to establish the result.
It's important that there must be a finite subcover for every open cover. Otherwise yes, trivially every set has a finite open cover; simply take it to be a single open set consisting of the union of open neighborhoods centered at every point in the set.
One way to think of the finite subcover condition is that, for a compact subset of a metric space, an open cover has to necessarily "overshoot" the set. In doing so, it becomes impossible for the minimal subcover to be infinite.
Example 1. Take $K = [0,1]$. If we have an open cover of $K$ by some open balls, then necessarily there is going to be an open ball that contains the element $1$. But if an open ball contains $1$, then that open ball must actually extend past $1$, in particular that open ball must contain all points greater than $1$ and less than $1 + \delta$ for some $\delta > 0$.
Example 2. Take $A = [0,1)$. This set is not compact, because we could choose the open cover consisting of balls $B_n = (-1, 1 - 1/n)$ for $n = 1, 2, \ldots$. This open cover never "overshoots" $A$ on the right endpoint; rather, the balls $B_n$ get ever and ever closer to $1$. Thus it is impossible for any subcover to be finite; we really do need to take a sequence with $n$ going off to infinity.
Best Answer
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.