In particular, what about the real line? If our topology is generated by sets of the form [a,b] or [a,b), why can't we form an open cover of, say, [0,1] with those and be guaranteed a finite subcover?
I'm sure it's a simple counterexample. I don't know how it's escaping me...
If the topology is generated by "open" sets of the form $[a,b]$, then singletons are also "open": $$[b-1,b]\cap[b,b+1] = \{b\}.$$ Hence, only finite sets can be compact in this topology.
The case $[a,b)$ is somewhat different: infite union of "open" sets has to be "open", hence we build an "open" set $$\bigcup_{n\ge 1}[a+1/n,a+2) = (a,a+2).$$ This allows us to conclude that this topology is finer than the usual one and therefore it has fewer compact sets. For example, the "open" cover of $[0,1]$ can be given by this model: let the sequence $a_n$ satisfy $$0=a_1,\quad \forall n\, a_n<a_{n+1},\quad \lim_{n\to\infty}a_n = 1.$$ We cover $[0,1]$ by "open" intervals $[a_n,a_{n+1})$, $n\in\Bbb N$ and one more interval $[1,2)$. Obviously, this cover has no finite subcovers, therefore $[0,1]$ is not a compact set in this topology.
One of problems in your approach that you should be able to find a finite subcover from any open cover, not just from a well-tailored cover.