Minimal axioms for topology (open sets definition) If $\langle X,\tau\rangle$ is a topological space implies $X=\bigcup\tau$, then the minimal axioms for topology can be stated as follows:

A topology $\tau$ is a set such that

*

*$\forall S\subseteq\tau.\bigcup S\in\tau$


*$\forall S\subseteq\tau.|S|<\aleph_0\implies\bigcap S\in \tau$

The notion of a "topological space" is essentially redundant, since the underlying set is implicit to the topology itself - that is, the underlying set of a topological space is uniquely determined by the topology. The rest of topology (the subject) can be formalised solely in terms of topologies (the objects), without ever referencing the idea of topological spaces.
The remaining axioms can be proven from $1$ (or $2$, if you prefer).

Theorem: If $\tau$ is a topology, then $\bigcup\tau\in\tau$.
proof: Let $\tau$ be a topology
$\begin{matrix}&\forall S\subseteq\tau.\bigcup S\in\tau & (\text{axiom 1})\\
&\tau\subseteq\tau & (\text{definition})\\
& \hline\therefore\bigcup\tau\in\tau& \ \end{matrix}$

...

Theorem: If $\tau$ is a topology, then $\emptyset\in\tau$
proof: Let $\tau$ be a topology
$\begin{matrix}&\forall S\subseteq\tau.\bigcup S\in\tau & (\text{axiom 1})\\
&\emptyset\subseteq\tau & (\text{definition})\\
& \hline\therefore\bigcup\emptyset\in\tau & \ \\\\&\bigcup\emptyset\in\tau&(\text{proven})\\&\bigcup\emptyset=\emptyset &(\text{definition})\\ &\hline\therefore \emptyset\in\tau\end{matrix}$

Is this sufficient to axiomatise topology? If so, why do we bother defining topological spaces at all?
Addendum
The answers given by Mirko and Theoretical Economist provide useful insight into why topology is thought of in terms of topological spaces rather than topologies.
I'm still unsure of whether or not the above axioms are 1) sufficient to axiomatise all of topology, 2) the smallest set of axioms for topology.
As pointed out by Henno Brandsma, the usual axioms of topology define a topological space as a pair $(X,\tau)$, with $\tau\subseteq\mathcal{P}(X)$, $X\in\tau$, and $\tau$ satisfying $(1)$ and $(2)$. Since it seems that $X=\bigcup\tau$ - unless there is a topological space $(X,\tau)$ s.t. $X\ne\bigcup\tau$ - it should suffice to say that a topological space is a pair $(\bigcup\tau,\tau)$, where $\tau$ satisfies $(1)$ and $(2)$.
Furthermore, it might be possible to prove $(1)$ and $(2)$ from a single statement (similar to how the Wolfram Axiom suffices to axiomatise Boolean algebra). I'm not yet sure how to derive such a statement, should one exist.
 A: It does look like what you've stated is sufficient to axiomatise topology. That said, I'm no expert on these issues, so there might be something I'm missing.
However, I can talk about your second question, which I'm interpreting as "why do we define topological spaces the way we do?"
One reason we do this is that it is often convenient in mathematics to treat sets as our primitives, and then consider various structures we can put on that set. This makes it natural to think of a topological space as consisting of a set and and a topology on that set.
For example, in measure theory, we are interested in measurable spaces, which are sets endowed with a $\sigma$-algebra. The $\sigma$-algebra is the collection of sets that you can "measure". Suppose you had some set $X$ which also has some topology $\tau$, then you might be interested in endowing $X$ with a $\sigma$-algebra that is, in some sense, compatible with your topology. This gives rise to the notion of the Borel $\sigma$-algebra, which is the smallest $\sigma$-algebra containing $\tau$.
Structures on sets that are compatible with each other are incredibly common in mathematics. Some examples come from the study of topological groups, topological vector spaces, topological manifolds, Borel $\sigma$-algebras, etc. Talking about these structures being compatible with each other is simplest when what you start with some underlying set with no structure, and then endowing that set with some structure, like a topology or a $\sigma$-algebra. You then impose some restrictions on one (or both) of these structures so that they are compatible in the way you need them to be.
A: Because in many(should I say most) cases you have the set with some structure on it already, and you impose additional structures on it, to better understand its properties. The set of reals is already there, with arithmetic operations, and then at some point you realize you want to study limits of sequences of real numbers, you don't want to dispose of the reals just because you could describe the topology without mentioning $X$ explicitly. Besides, it is VERY convenient to be able to talk about the elements $x$ of $X$, in various proofs and constructions: You do not want to talk just about open sets, i..e just about $\tau$ and its elements. Some redundancy in the description is beneficial, to keep things tidy. 
A: If you want a ZFC first order description of a topology:
A topology is a pair of sets $(X,\tau)$ such that


*

*$\forall A \in \tau: \forall x \in A: x \in X$ (or in abbreviated notation $\tau \subseteq \mathscr{P}(X)$.

*$X \in \tau$.

*$\forall \tau': (\forall A \in \tau': A \in \tau) \to \bigcup \tau' \in \tau$ or in abbreviated notation: $\forall \tau': (\tau' \subseteq \tau) \to \bigcup \tau' \in \tau$, which does imply the usual $\emptyset \in \tau$ axiom by using empty unions.

*$\forall A,B: (A \in \tau \land B \in \tau) \to A \cap B \in \tau$ (using "finite" is awkward as this is a derived, not a primimitive first order notion), and in the metamathematics finite induction is allowed to show finite intersections.
You do have to assume $X \in \tau$ somewhere, as otherwise you get different, unintended models.
