Your definition of compactness is wrong: a space is compact if any family of closed subsets has the property that if the intersection of any finite subfamily is non-empty, then the whole intersection is non-empty. The finite and closed (not any, as you suggest by subset of its powerset) part are essential: a simple example : take all subsets of $[0,1]$ (that contain a measurable set) of Lebesgue measure at least $\frac34$: any two of them must intersect, but their total intersection is empty; and this while $[0,1]$ is actually compact.
Another closure-like characterisation of compactness: recall that a closure point $x$ of a set $A$ is a point such that every open neighbourhood of $x$ intersects $A$. A strengthening is a point of total accumulation of an infinite set $A$: $x$ is called a point of total accumulation of $A$ when every open neighbourhood of $x$ intersects $A$ in a set as large as $A$ itself, i.e. $$\forall O \text{ open }: x \in O \implies |O \cap A| = |A|$$ where $|B|$ denotes the cardinality of a set $B$.
Theorem: A space $X$ is compact iff every infinite subset $A$ of $X$ has a point of total accumulation in $X$.
This is a closure-like property that might give you another way to think about compactness.
Separability is just that there exists a countable set $D$ such that $\overline{D} =X$.
First-countability does not really have a closure-like reformulation that I can think of. Neither does second-countability or $\sigma$-compactness. For linearly-Lindelöfness there is a reformulation in terms of special closure points as well, but it's more technical, but not for Lindelöf.
For Lindelöf we do have the analogon of the finite intersection property formulation:
$X$ is Lindelöf iff every family of closed subsets of $X$ that has the countable intersection property, has non-empty intersection.
