Show that $\tau\left(\frac{1}{n}\right)$ defined as $\tau\left(\frac{1}{n}\right)=\{A\subset\mathbb{R}: (\exists N\in\mathbb{N}) \forall(n\in\mathbb{N},n\geq N)\frac{1}{n}\in A\}\cup\{\emptyset\}$ is a topology for $\mathbb{R}$.
$\textbf{Proof:}$ In order to to prove this family of subsets is a topology, it must satisfy 3 conditions.
$i)$ $\emptyset,X\in\tau$.
$ii)$ If $A_1,A_2\in\tau$ then $A_1\cap A_2\in\tau$.
$iii)$ If $\{A_\alpha\}_{\alpha\in a}$ are elements of $\tau$ then $\displaystyle\bigcup_{\alpha\in a}{A_\alpha}\in\tau$.
We will show all three.
i) Clearly $\emptyset\in\tau$. Since $\mathbb{R}\ni\frac{1}{2},\frac{1}{3},\frac{1}{4}...$ then $\exists N\in\mathbb{N}$ (namely $N=2),\forall n\geq N,\frac{1}{n}\in\mathbb{R}$
Therefore $\emptyset,X\in\tau(\frac{1}{n})$ and thus condition $i$ is satisfied.
$ii)$ Assume $A_1,A_2\in\tau(\frac{1}{n})$.
If either $A_1=\emptyset$ or $A_2=\emptyset$ then $A_1\cap A_2=\emptyset\in\tau(\frac{1}{n})$ and we are done.
Otherwise assume $A_1\neq\emptyset$ and $A_2\neq\emptyset$ and $n_1\leq n_2$.
$A_1\in\tau(\frac{1}{n})\Rightarrow A_1\subset\mathbb{R}$ and $A_1\ni\frac{1}{n_1},\frac{1}{n_1+1},\frac{1}{n_1+2},...,\frac{1}{n_1+m}$ where $m\in\mathbb{N}$.
$A_2\in\tau(\frac{1}{n})\Rightarrow A_2\subset\mathbb{R}$ and $A_2\ni\frac{1}{n_2},\frac{1}{n_2+1},\frac{1}{n_2+2},...,\frac{1}{n_2+m}$ where $m\in\mathbb{N}$.
Because $n_1,n_2\in\mathbb{N}$ and $n_1\leq n_2$ then $A_1\cap A_2\ni\frac{1}{n_2},\frac{1}{n_2+1},...$.
Thus $A_1\cap A_2\subset\mathbb{R}\in\tau(\frac{1}{n})$.
Therefore $A_1\cap A_2\in\tau(\frac{1}{n})$ and condition $ii)$ is satisfied.
This is what I have so far. Please correct me if I'm wrong up to this point. I'm stuck trying to prove the last condition. Some help would be greatly appreciated. Please and thanks.
What you have done seemed OK to me.
For the union, let $(A_i)_{i\in I}$ be a family of elements of $\tau(1/n)$.
For each $i \in I$, there exists $N_i \in \mathbb N$ such that $$n \geq N_i \Longrightarrow 1/n \in A_i.$$ Now $\{N_i:i\in I\}$ is non-empty subset of $\mathbb N$, so it has a minimum element, $N_{i_0}$, that is, there exists $i_0 \in I$ such that $$N_{i_0} = \min\{N_i:i\in I\}.$$ It follows that $$A_{i_0}=\bigcup_{i\in I}A_i.$$ So not only the union of a family is an open set but it is also a member of that family.