Do we need finite intersections of open sets to be open? How important/indispensable for the general theory of topology is the requirement that the open sets are closed under finite intersections?
If I am not overlooking something crucial, then a huge part of mainstream topology remains true if "topology" is replaced by "generalized topology", which only requires unions of open sets to be open. This seems to be true for the main topological pillars, i.e. continuity and compactness up to Tychonoff's theorem. If this is indeed so, why is 95% of literature placing so much emphasis on the finite intersection requirement? Which important topological theorems fail without it?
 A: There is an issue with products. Consider the following standard fact from ordinary topology.
Proposition: If $f:X\to A$ and $g:X\to B$ are continuous, then $(f,g):X\to A\times B$ is continuous.
Proof: Let $U\subset A$ and $V\subset B$ be open subsets. Then $(f,g)^{-1}(U\times V)=f^{-1}(U)\cap g^{-1}(V)\subset X$ is open. The sets $U\times V$ are a basis for the topology on $A\times B$, so $(f,g)$ is continuous. $\blacksquare$
The proof needs the intersection of two open subsets of $X$ to be open, so fails if $X$ is a generalized topological space. 
It turns out that the product topology is the wrong one for working with generalized topologies. Anything we call a product $A\times B$ ought to have the property that a map $(f,g):X\to A\times B$ is continuous if and only if $f$ and $g$ are continuous. There is a generalized topology with this property: a subset $C\subset A\times B$ is open iff for every $(a,b)\in C$, either there exists an open neighborhood $U\subset A$ of $a$ with $U\times B\subset C$, or there exists an open neighborhood $V\subset B$ of $b$ with $A\times V\subset C$.
The generalized product topology is what it is, but there are some problems. Consider another fact from ordinary topology.
Proposition: Suppose $f,g:X\to \mathbb{R}$ are continuous functions. Then $f+g$ is continuous.
Proof: The functions $(f,g):X\to\mathbb{R}^2$ and $a:\mathbb{R}^2\to \mathbb{R}$, $(x,y)\mapsto x+y$ are continuous, so the composition $a\circ (f,g)=f+g$ is also continous. $\blacksquare$
This fails if $X$ is a generalized topological space: if we take the ordinary product topology on $\mathbb{R}^2$ then $(f,g)$ might not be continuous, and if we take the generalized product topology on $\mathbb{R}^2$ then $a$ is not continuous. The counterexample I gave in the comments is $X=\{1,2,3\}$ where we define every subset except $\{1\}$ to be open, $f$ is the characteristic function of $\{2\}$, and $g$ is the characteristic function of $\{3\}$. Then $f$ and $g$ are continuous, but $f+g$ is not.
