Let $X$ be a topological space. Show that if every filter $F$ on $X$ converges to at most one point, then $X$ is Hausdorff.
I want to argue by contrapositive. Suppose there exists a pair of points $x$ and $y$ such that every pair of their neighborhoods intersect. Then I want to construct a filter that consists of all their neighborhoods and hence converges to both $x$ and $y$. I haven't been able to construct such a filter. Is this the correct way to proceed or are there any better methods?
The contrapositive is the right idea. The neighbourhood filters are key:
If $x$ and $y$ are such points let $\mathcal{B}=\{U \cap V: U, V \text{ open }, x \in U, y \in V\}$, which by assumption only consists of non-empty sets.
$\mathcal{B}$ forms a filter base in $X$ so it generates a filter $\mathcal{F} = \{A: \exists B \in \mathcal{B}: B \subseteq A\}$ on $X$ such that $\mathcal{F} \to x$ and $\mathcal{F} \to y$. This is trivial, as all neighbourhoods of $x$ and $y$ will be in $\mathcal{F}$.