Let $(\mathbb R, T_{CF})$. Let $s=(s_d)_{d\in D}$ a net in $\mathbb R$. Proof that: $$\mathcal Lim(s) =\mathbb R\iff \bigcap_{d\in D}B_d=\emptyset$$
Notation
$T_{CF}=\{\emptyset, \mathbb R\}\cup \{A\subset \mathbb R\mid \mathbb R\setminus A \text { is finite}\}$, the co-finite topolgy on $\mathbb R$.
$B_{d_0}=\{s_d\mid d\geq d_0\}\ \forall d_0\in D $.
$x\in \mathcal L im (s)$ by definition if $\forall U$ neighbourhood of $x$ in $(\mathbb R,T_{CF})$, $\exists d_0\in D$ such that $s_d\in U\ \forall d\geq d_0$.
My attempt
I have some conclusions but I don't get the proof:
$\Rightarrow$) As $\mathcal Lim(s)\subset agl(s)$ and $agl(s)=\bigcap_{d\in D}\overline {B_{d}}$, we have that $\mathbb R =\overline{B_d}\ \forall d\in D$. In $(\mathbb R,T_{CF})$, the closed sets are $\mathcal C_{T_{CF}}=\{\emptyset, \mathbb R\}\cup \{F\subset \mathbb R\mid F \text { is finite}\}$. So we conclude that $B_d$ must be infinite $\forall d\in D$ (If it were finite, will be closed and its clousure would coincide with itself). I don't know how to proceed after that.
I have tried also to prove this implication with the contaposition argument: If $\bigcap_{d\in D}{B_{d}}\neq \emptyset$, then $\exists s_{d_0}\in B_d\ \forall d\in D$. But I don't know how to continue this argument to prove that $\mathcal L im (s)\neq \mathbb R$.
$\Leftarrow)$ By contraposition: If $\mathcal L im (s)\neq \mathbb R$, then $\exists x_0\in \mathbb R$ that $s$ doesn't converge to $x_0$. This is that $\exists U$ neighbourhood of $x_0$ such that$ \forall d_0\in D \ \exists d\in D$ with $d\geq d_0$ and $s_d\notin U$. So $B_d\not\subset U \ \forall d\in D$. But, again, I don't know how to get a proof with this.
Please, if you can guiude me I would appreciate that.
Suppose $\mathcal{Lim}(s)=\Bbb R$. Suppose we have $\bigcap_{d \in D} B_d \neq \emptyset$, pick $p \in \Bbb R$ in it. This means that for each $d \in D$ we have $p \in B_d$ so that there is some $d' \ge d$ so that $s_{d'}=p$. So we have a cofinal set on which the net only assumes the value $p$. If now $q \neq p$ is another point of $\Bbb R$, then, by definition of the cofinal topology, $U = \Bbb R - \{p\}$ is an open neighbourhood of $q$ and by assumption $q$ is a limit of the net (all points of $\Bbb R$ are) and so for some $d_1 \in D$ we have that $\forall d \ge d_1: s_d \in U$ so in particular $s_d \neq p$. So $p \notin B_{d_1}$, contradicting how $p$ was chosen. Note that we only used the topology is $T_1$ for this direction.
Now suppose $\bigcap_{d \in D} B_d = \emptyset$. Let $x \in \Bbb R$; we have to show it's a limit of the net, so let $U$ be an open neighbourhood of $x$, so $U= \Bbb R-F$ where $F$ is a finite subset of $\Bbb R$. For each $f \in F$ we know that $f \notin \bigcap_{d \in D} B_d$ so there must be some $d(f) \in D$ so that $f \notin B_{d(f)}$.
Having these finitely many $d(f) \in D$, because we have a directed set, we can find $d_0 \in D$ so that $d_0 \ge d(f)$ for all $d(f), f \in F$.
So when $d \ge d_0$, $d \in B_{d(f)} \not\ni f$ for all $f \in F$, and this implies $s_d \in \Bbb R-F = U$. As $U$ was arbitrary, indeed $x \in \mathcal{Lim}(s)$ and as $x$ was arbitrary, $\mathcal{Lim}(s)=\Bbb R$ and we're done.