Let $\{x_i\}_{i\in I}$ be a net in a topological space. Define a filter base $\mathcal{F}$ such that for all $x$, $\mathcal{F}\rightarrow x$ if and only if $x_i\rightarrow x$.
My efforts: Define $\mathcal{F}=\{N:N\text{ is a neighborhood of }x_i\text{, }i\in I\}$.
One direction is easy: If $x_i\rightarrow x$, then given any neighborhood $N$ of $x$, there exists $\alpha\in I$ such that $x_\beta\in N$ if $\beta\geq\alpha$. Therefore, $N$ is a neighborhood of $x_\beta$ and thus $N\in\mathcal{F}$.
The other direction fails. For example, $\{x_i\}_{i\in I}$ has two subnets, one converges to $x$ and the other converges to $y$. This case can also lead to $\mathcal{F}\rightarrow x$.
The filter of tails $\mathcal{F}$ is generated by the filterbase $\mathcal{B}:=\{T(i): i \in I\}$ where $T(i)= \{x_j: j \in I, j \ge i\}$ for all $i \in I$.
If $\mathcal{F} \to x$ then let $N$ be a neighbourhood of $x$. Then $N \in \mathcal{F}$ so there is some $T(i) \in \mathcal{B}\}$ such that $T(i) \subseteq N$. But this means that $\forall j \ge i: x_j \in N$ by definition. So $x_i \to x$.
The reverse is almost the same idea: if $x_i \to x$, let $N$ be a neighbourhood of $x$, so we know there is some $i_0 \in I$ such that all $x_i \in I$ with $i \ge i_0$ are in $N$. This just says that $T(i_0) \subseteq N$ and so by definition $N \in \mathcal{F}$. As $N$ was arbitrary, $\mathcal{F} \to x$.