Regarding the definition in terms of open covers, this turns out to be technically very useful, but I agree that it can be hard to understand at first. You might want to consider one of the equivalent definitions in terms of closed sets.

The contrapositive of the standard open cover condition says that if $U_i$ is a
collection of open sets so that no finite union of $U_i$'s covers the compact set
$X$, then the entire set of $U_i$'s doesn't cover $X$. Passing to complements, we find that if $F_i$ is a collection of closed subsets having non-empty intersection with $X$, then the intersection of the entire collection has non-empty intersection with $X$.

Now this can be rephrased in a slightly simpler way: given a chain of closed subsets (i.e. a collection of non-empty closed subsets such that for any two members of the collection, one is contained in the other) each having non-empty
intersection with $X$, their intersection has non-empty intersection with $X$. (For a discussion of this, see this answer, or more-or-less any topology text book.)

So this says that we can't find a shrinking collection of closed subsets each having a point in common with $X$, but whose intersection doesn't have a point in common with $X$.

Contrast with say the interval $X := (0,1).$ The subsets $[0,1/n]$ are closed intervals having non-empty intersection with $X$, but their intersection is the
singleton $\{0\}$, which doesn't have non-empty intersection with $X$.

Consider also the whole real line. The subsets $[n,\infty)$ are closed and non-empty, but their intersection is empty.

So these two examples capture the basic intuition for how compactness fails: a set may be missing some kind of "boundary point" or "point at infinity". Conversely, the intuition is that a compact set is not missing any such points.

The definition in terms of subsequences necessarily containing a convergent subsequence (valid in a metric space) captures the same idea.

As in the above linked answer, one can also rephrase the definition in terms of chains of open sets. Then it says that if $U_i$ is any chain of open sets whose union covers $X$, then already one of the of open sets $U_i$ must contain $X$.

So you can think of this as saying that it is not possible to fill out $X$ by taking an increasing union of proper open sets (in the induced topology).

Compare this with the open interval $(0,1)$, which is the increasing union of the proper open subsets $(1/n,1-1/n)$.

Also, here is an aside on index sets, which can sometimes be confusing when you first encounter them: note that while one can't reduce to countable open covers, or decreasing sequences of closed subsets, in the general definition of compactness (and one can't use sequences to test for compactness in general --- although there is
a generalization of sequences (in fact more than one --- *nets* or *filters*) which can be used), when building up intuition it is fairly safe to imagine that the open covers in question are countable, or that the chain of closed subsets is just a decreasing sequence of closed subsets.

While it is technically important to learn to argue with non-countable index sets as in the general definition of compactness, I wouldn't let the possible non-countability of those index sets be the thing I focus on when building up intuition; this truly is a technical point which will just get in the way from
an intuitive view-point. As you become more confident in manipulating index sets, the distinction between the countable and general case won't seem like that big of a deal anyway.