Exercise about convergence of a net in $(\mathbb R, T_{CF})$ Let $(\mathbb R, T_{CF})$ topological space with $T_{CF}=\{U\subset \mathbb R\mid \mathbb R\setminus U \text{ is finite}\}$ the co-finite topology. Let $s=(s_d)_{d\in D}$ a net in $\mathbb R$ such that $\bigcap_{d\in D}B_d\neq\emptyset$ with $B_{d_0}=\{s_d\in \mathbb R\mid d\geq d_0\}$. Let $x_0\in \mathbb R$. Then:
$$x_0\in\mathcal Lim(s)\iff \bigcap_{d\in D}B_d=\{x_0\}$$
My attempt
$\Rightarrow)$ $\subset)$ Proof by contradiction: Supouse that $\exists x_1\in \bigcap_{d\in D}B_d$ with $x_1\neq x_0$. Then $\mathbb R\setminus\{x_1\}$ is an open neighbourhood of $x_0$ in $(\mathbb R, T_{CF})$. As, by hypothesis, $x_0$ is a limit point of $s$, $\exists d_0 \in D$ such that $s_d\in\mathbb R\setminus \{x_1\}\ \forall d\geq d_0$. Then, $B_{d_0}\subset \mathbb R\setminus\{x_1\}$. Therefore $x_1\notin B_{d_0}$, so $x_1\notin \bigcap_{d\in D}B_d$ !! Contradiction
$\supset)$ Proof by contradiction: Supouse that $\exists x_0\notin \bigcap_{d\in D}B_d$. As $\bigcap_{d\in D}B_d\neq \emptyset$, $\exists y\in \bigcap_{d\in D}B_d$ with $y\neq x_0$. Then $\mathbb R\setminus\{y\}$ is an open neighbourhood of $x_0$ in $(\mathbb R, T_{CF})$. As, by hypothesis, $x_0$ is a limit point of $s$, $\exists d_0 \in D$ such that $s_d\in\mathbb R\setminus \{y\}\ \forall d\geq d_0$. Then, $B_{d_0}\subset \mathbb R\setminus\{y\}$. Therefore $y\notin B_{d_0}$, so $y\notin \bigcap_{d\in D}B_d$ !! Contradiction
$\Leftarrow)$ Proof by contradiction: Supouse that $x_0$ is not a limit point of $s$. Then $\exists U=\mathbb R\setminus \{x_1,...,x_n\}$ an open neighbourhood of $x_0$ such that $\forall d_0\in D\ \exists d\geq d_0$ such that $s_d\notin U$. So $s_d\in\{x_1,...,x_n\}$.
I don't know how to conclude the sufficent condition and I also don't konw if there is another way to prove the necesary condition easier than mine. If youy could guide me, I appreciate that.
 A: Fix some $p \in \Bbb R$ in $\bigcap_{d \in D} B_d$, to use the assumption on the net. It follows that $D'=\{d \in D\mid s_d = p\}$ is cofinal in $D$.
For the forward direction, if $x_0 \in \mathcal{L}im(s)$, it must be the case that $x_0=p$, because otherwise $\Bbb R\setminus \{p\}$ would be an open neighbourhood of $x_0$ and would contain all $s_d$ from some $d_0$ onwards, contradicting the cofinalness of $D'$. It follows that in these circumstances (i.e. there being a limit $x_0$ at all and $X$ being $T_1$) $\bigcap_{d \in D} B_d$ cannot have more than one element, so we have the right hand condition: one element, the limit.
Backward: suppose $\{x_0\} = \bigcap_{d \in D}  B_d$. Let $U$ be any open neighbourhood of $x_0$, so $U=\Bbb R\setminus F$ where $F$ is finite and disjoint from $x_0$. So for any $f \in F$ we have that $f$ is not in the intersection $\bigcap_{d \in D} B_d$ and so for some $d_f$, $f \notin B_{d_f}$, so $$\forall d \ge d_f: s_d \neq f$$ and as $D$ is directed we can find one $d_1 \in D$ such that $$\forall f \in F: d_1 \ge d_f$$ and so
$$\forall d \ge d_1: s_d \notin F$$ and so
$$\forall d \ge d_1 : s_d \in U$$
and as $U$ was an arbitrary open neighbourhood of $x_0$ we have $x_0 \in \mathcal{L}im(s)$, as required.
