Summary
"open (or closed) topology on $X$"= "join-complete (meet-complete, respectively) bounded lattice on $\mathcal P(X)$, ordered by inclusion."
"theory of topology"="theory of join-complete (meet-complete) bounded lattices."
Complements are optional for the "theory of join-complete (meet-complete) bounded lattices," so "closed sets" (open sets) should be optional for the "theory of topology."
Do we need closed (or open) sets? If not, how do we describe topological properties exclusively in terms of open (closed) sets? i.e. what is "boundary," "connected," "interior," etc. in terms of open (or closed) sets alone?
1. Theory of Topology vs Topological Spaces
In my experience, topology is usually defined as the study of topological spaces which are in turn defined to be specific kinds of sets ("with additional structure," but unless we want to talk about topologies on proper classes a "structure" is still a kind of set.) I always thought that this was weird because it makes topological properties dependent on the specific choice of set theory - a point which is usually glossed over in "real" topology.
Put differently, if we fix a set theory $\mathsf{ST}$ (with some additional caveats) and let $\mathsf{TS}$ be the the theory obtained from the axioms of topology - whatever those may be - then either $\mathsf{TS}\subseteq\mathsf{ST}$ or $\mathsf{TS}$ is an extension-by-definitions of $\mathsf{ST}$. Knowing this, it's hard to justify the position that topology is anything other than set theory itself. I mean, if we want to be really obnoxious about it I guess we could make the claim that the symbols '$\cap$,' '$\cup$,' etc. (or the words 'intersection,' 'union,' etc., for that matter - all symbols in the general sense) which might appear in the axioms of topology do not belong to $\mathsf{ST}\subseteq\langle\in\rangle$, and thus $\mathsf{TS}$ truly constitutes a separate theory from $\mathsf{ST}$. But is anyone really claiming that 'union' in topology doesn't refer to the union of sets?
This is made more unusual by the fact that it isn't particularly difficult to create a "proper" theory of topology. For example, we have the second-order theory - which I'll call $\mathsf{LT_2}$ - in the language $\langle\wedge,\vee\rangle$ given by the axioms:
$\exists x\forall y((x\wedge y=x)\land(x\vee y=y))$, i.e. the empty set exists (is open)
$\exists x\forall y((x\wedge y=y)\land(x\vee y=x))$, i.e. the universal set exists (is open)*
$\forall X\exists x\forall y(X(y)\implies((x\wedge y=y)\land (x\vee y=x)))$, i.e. the arbitrary union of sets exists (is open)
*this axiom is redundant
(note that closure under intersection is guaranteed by the fact that $\wedge$ is a binary function symbol, hence $x\wedge y$ is a term and $\forall x\forall y\exists!z(x\wedge y=z)$ is a [trivial] theorem)
The standard models ("lattice topologies") are obtained from the obvious interpretation $\wedge\to\cap,\vee\to\cup$ with the domain being a topology (the set of open sets) on the underlying set of the corresponding topological space. For example, if
$$E=\left\{\bigcup_{S\in I}S:I\subseteq\{(\alpha,\beta)\subseteq\Bbb R:\alpha\le\beta\}\right\}$$
Then $(E,\cap,\cup)\vDash\mathsf{LT_2}$ (it is, in fact, the standard topology on $\Bbb R$ dressed as a lattice.) These models aren't isomorphic to conventional topological spaces but we can construct something akin to isomorphism by way of functions $F_X:\{(\tau_X,\cap,\cup):\bigcup\tau_X=X, (\tau_X,\cap,\cup)\vDash\mathsf{LT_2}\}\to\mathbf{Top}(X)$, where $\mathbf{Top}(X)$ is the set of all topological spaces on $X$, such that:
$$F_X(\tau_X,\cap,\cup)=(X,\tau_X)$$
The union of all such functions is a class function [also a functor, I think?] which maps lattice topologies to topological spaces.
This may or may not happen to be the way that pointless topology is done - I wouldn't know; but it doesn't really matter since the whole point of this was to show that we can create a theory of topology with minimal effort.
2. Topology vs Closed (or Open) Sets
While we can show that $\mathsf{LT_2}$ (or any other theory) is essentially the "theory of topology" there is one noteworthy distinction. In $\mathsf{TS}$ we can prove the existence ($\pm$uniqueness) of any set whose existence we can prove in $\mathsf{ST}$. In particular - if $\mathsf{ST}$ has [relative] complements, as is the case for $\mathsf{ST}=\mathsf{ZFC}$ - we can define closed sets without departing from $\mathsf{TS}$ (actually, this isn't true, but you have to go out of your way to make sure that $\mathsf{TS}$ doesn't admit complements.)
This is not the case for $\mathsf{LT_2}$, where the only sets that exist are open sets. We can readily see this by looking at finite topologies. Consider, for example, the set
$$A=\{\emptyset,\{a\},\{c\},\{a,c\},\{a,b,c\}\}$$
Clearly $\mathcal A=(A,\cap,\cup)$ is a structure for $\mathsf{LT_2}$, but the closed set $\{b\}$ is excluded from the domain. Of course, we can still talk about $\{b\}$ as an object of $\mathcal A$ within $\mathsf{TS}$, but we can infer nothing about its behaviour from $\mathsf{LT_2}$.
This, together with the fact that $\mathsf{LT_2}$ suffices to characterize all topological spaces (to the extent that any topological space is characterized by $\mathsf{TS}$) leads me to question whether or not closed sets (or open sets, if we use a closed set definition) were even necessary in the first place.
It is not hard to add closed sets to $\mathsf{LT_2}$ (or any other theory) - simply add complements and the predicates '$O$' and '$C$' with the axiom $\forall x(O(x)\iff C(x'))$, where $x'$ is the complement of $x$. But it would be much more interesting to find out that closed sets (or open sets) are redundant.