I have been thinking about this one for awhile, but I cannot crack it. I think the proof to this is similar to the proof that the topology containing all singleton sets is the discrete topology, in that that proof used the infinite union of singletons to build every subset. In this one I imagine that some how the use of finite intersections can create every finite subset of $X$, but that seems weird and I'm not sure how I would show that.
Is this correct? Can someone show me where to start?
Hint. There is an injection $i:\Bbb N\to X$.
For $x\in X$ consider the infinite sets $i(2\Bbb N)\cup \{x\}$ and $i(2\Bbb N+1)\cup \{x\}$.
I'm assuming that $X$ is infinite, otherwise the claim does not hold.
If suffices to prove that the singletons $\{x\}$ are open, for all $x \in X$.
$X \setminus \{x\}$ is also an infinite set so there exist infinte sets $A, B \subseteq X \setminus \{x\}$ such that $A \cap B = \emptyset$.
For example, let $\{a_n : n \in \mathbb{N}\}$ be a countable subset of $X \setminus \{x\}$ and then define $A = \{a_{2n} : n \in \mathbb{N}\}$ and $B = \{a_{2n - 1} : n \in \mathbb{N}\}$.
Now $A \cup \{x\}$ and $B \cup \{x\}$ are infinite subsets of $X$ so they are open.
Hence $$\{x\} = (A \cup \{ x \}) \cap (B \cup \{ x \})$$ is also open as an intersection of two open sets.
Now an arbitrary $S \subseteq X$ is just
$$S = \bigcup_{x \in S} \{x\}$$
which is a union of open sets.
This is a good start. Definitely take advantage of the fact that the intersection of any two open sets in a topological space is itself an open set. In particular, think about$^\dagger$ how you can get an arbitrary singleton set $\{x\}$ by an intersection of two sets known to be open in this space.
Demonstrating that an arbitrary singleton set is open implies that all of them are. Because the set of all singleton sets is a base for the discrete topology, this shows that $\tau$ is indeed discrete$^{**}$. In other words, notice that you can now write any subset $Y \subset X$ as a union of sets shown to be open:
$$Y = \bigcup_{y \ \in \ Y} \ \{y \}$$
$^\dagger$Regardless of the cardinality of $X$, we can always find an injection $\phi: \mathbb{N} \hookrightarrow X$. If we wish to show that $\{x\}$ is an open set, it may be convenient to construct a $\phi$ in such a way as to have $x \in \phi(\mathbb{N})$. This way, you can first do the problem in the natural numbers, trying to get the singleton $\big\{\phi^{-1}(x) \big\}$ as an intersection of two infinite subsets of $\mathbb{N}$; the advantage is that $\mathbb{N}$ has labeled elements and an ordering for ease of thought. When done, the idea can be transferred to $X$ via $\phi$ (that such a $\phi$ is guaranteed to exist is sufficient).
$^{**}$This is a common theme in topology. Often, you can show that something is true about a topological space as long as you can show that it's true about a base for the topology in question. For instance, the fact used above was: if $\mathcal{B}$ is a base for a topology $\tau'$, then $\mathcal{B} \subset \tau \implies \tau' \subset \tau$. Another that quickly comes to mind is that a function $f:X \rightarrow Y$ between topological spaces is an open map as long as the function is an open map with respect to the elements in a base for $X$, due to the fact that $\displaystyle f \left( \bigcup_k U_k \right) = \bigcup_k f(U_k)$. In the same vein, it is sufficient, when proving that a function is continuous, to show only that the preimages of $Y$'s base elements are open in $X$.
Consider a point $x\in X$ and choose a partition of $\{x\}^c$ as $\{x\}^c=A \coprod B$, where both $A$ and $B$ infinite.
Then $A\coprod \{x\},B\coprod \{x\} \in \tau$ and therefore $\{x\}=(A\coprod \{x\})\cap (B\coprod \{x\}) \in \tau$.
Hence each $\{x\}\in \tau$ is open and therefore $\tau$ is discrete.
