Let $X$ be an infinite set. Show that the intersection of all Hausdorff topologies on $X$ is a topology, and find it (Hint: We are familiar with this topology).
The best candidate seems to be the co-finite topology on X, as it is a subset of each of the Hausdorff topologies, and it fits the hint. However, I am having trouble proving that for every subset of $X$ whose complement is infinite, there is a Hausdorff topology which does not include it.
You want to prove that $$\bigcap T_2(X) = \tau_{cofin}(X),$$ where $T_2(X)$ is the collection of $T_2$ topologies on $X$ and $\tau_{cofin}(X)$ is the co-finite topology on $X$.
It seems that you already noticed that the reverse inclusion is true.
So, given a subset $A$ of $X$ such that $X \setminus A$ is infinite, you want to find a Hausdorff topology on $X$ such that $A$ is not open in that topology.
So fix $a_0 \in A$ and define $\tau_0$ to be the family of subsets of $X$ defined by:
$$U \in \tau_0 \;\Leftrightarrow\; \begin{cases} a_0 \notin U \;\text{ or}\\ a_0 \in U \;\text{ and } U \text{ is co-finite.} \end{cases} $$ Clearly, $A \notin \tau_0$.
Let us check that $\tau_0$ is a topology on $X$ (on which $A$ is not open).
Let $U,V \in \tau_0$. If $a_0 \notin U$ or $a_0 \notin V$, then $a_0 \notin U \cap V$. If $a_0 \in U$ and $a_0 \in V$, then $U$ and $V$ are co-finite, and then so is $U \cap V$. So $\tau_0$ is closed under finite intersections.
Let $(U_i)_{i \in I}$ be an arbitrary sub-family of elements of $\tau_0$. If $a_0 \notin U_i$, for all $i \in I$, then $a_0 \notin \bigcup_{i \in I}U_i$, whence $\bigcup_{i \in I}U_i \in \tau_0$. If there exists $i_0 \in I$ such that $a_0 \in U_{i_0}$, then $U_{i_0}$ is co-finite, whence $$X \setminus \bigcup_{i \in I} U_i = \bigcap_{i \in I} X \setminus U_i \subseteq X \setminus U_{i_0},$$ and so $\bigcup_{i \in I} U_i$ is co-finite. Hence $\tau_0$ is closed under arbitrary unions.
Clearly $\varnothing, X \in \tau_0$ and we conclude that $\tau_0$ is a topology on $X$.
Let us see that $\tau_0$ is Hausdorff.
Let $x,y$ be two different elements of $X$.
We may, without loss of generality, suppose that $y \neq a_0$.
Let $U_x = X \setminus \{y\}$ and $U_y = \{y\}$.
Then $x \in U_x$ and $y \in U_y$; and $U_x \cap U_y = \varnothing$.
And of course, $U_y \in \tau_0$ because $a_0 \notin U_y$, and $U_x \in \tau_0$ because $U_x$ is co-finite.