I don't know if there's still interest in this thread, but I'm going to answer anyway.
As Asaf points out, $\bigcap \emptyset$ may not exist, and it certainly doesn't equal $X$. The problem occurs because subsets don't remember the larger set they were cut from. However, this is a quirk of how subsets are defined. And, the problem can be avoided by redefining the meaning of "subset." This will allow us to rewrite the definition of a topological space.
Firstly, lets agree that by "function", we will mean an ordered triple $(f,X,Y)$. Then we can define that a subset of $X$ is a function $A : X \rightarrow 2,$ where $2 = \{0,1\}.$
Now for a bit of notation. As shorthand for $A(x)=1$, lets write $x \propto A,$ which can be read "$x$ is an element of $A$."
Lets also write $A \diamond X$ to mean that $A$ is a subset of $X$. Note that the subset relation is no longer transitive, so we also need a containment relation. So if $A,B \diamond X$, lets write $A \subseteq B$ in order to mean that for all $x \in X$ it holds that if $x \propto A$, then $x \propto B$.
Also, lets write $2^X$ for the powerset of $X$. Formally, we define $2^X := \{A : X \rightarrow 2 | A \mbox{ is a function}\}.$
Thus $A \diamond X$ if and only if $A \in 2^X$.
Finally, lets write $A = \{x \in X | P(x)\}$ in order to mean that $A \diamond X$, and that $x \propto A$ iff $P(x)$.
Given these conventions, we can define the intersection of $\mathcal{A} \diamond 2^X$ as follows.
$\bigcap \mathcal{A} = \{x \in X|\forall A \propto \mathcal{A} : x \propto A\}$
Unions can be defined similarly.
Finally, the payoff: letting $\bot$ denote the least subset of $2^X$, and letting $\top$ denote the greatest subset of $X$, we see that
$\bigcap \bot = \top.$
We're now in a position to rewrite the definition of a topological space.
A set $X$ with a subset $\tau \diamond 2^X$ is called a topological space if:
For all $\mathcal{A} \subseteq \tau$ it holds that $\bigcup \mathcal{A} \propto \tau$.
For all finite $\mathcal{B} \subseteq \tau$ it holds that $\bigcap \mathcal{B} \propto \tau$.