Is the following argument correct?
Proposition. $\tau_3$ consisting of $\mathbf{R},\varnothing$ and every interval $[n,\infty)$, for $n$ any positive integer, is a topology on $\mathbf{R}$.
Proof. Let $\mathcal{A}$ be an arbitrary collection of elements of $\tau_3$. In addition we may assume that $\varnothing\not\in\mathcal{A}$. We prove that $\bigcup\mathcal{A}\in\tau_3$. If $\mathbf{R}\in\mathcal{A}$, then $\bigcup\mathcal{A} = \mathbf{R}$, since $X\subseteq\mathbf{R},\forall X\in\tau_3$. So now assume that $\mathbf{R}\not\in\mathcal{A}$, consequently $\mathcal{A}\subseteq\{[n,\infty):n\in\mathbf{Z}^+\}$.
Now either $\mathcal{A} = \varnothing$ in which case $\bigcup\mathcal{A} = \varnothing\in\tau_3$, but if $\mathcal{A}\neq\varnothing$, then the set $\Omega = \{n\in\mathbf{Z}^+:[n,\infty)\in\mathcal{A}\}$, is non-empty and by the well-ordering principal has a least element $\alpha$, consequently $[x,\infty)\subseteq[\alpha,\infty),\forall [x,\infty)\in\mathcal{A}$ and by extension $\bigcup\mathcal{A} = [\alpha,\infty)\in\tau_3$.
Now let $X_1,X_2\in\tau_3$. We prove that $X_1\cap X_2\in\tau_3$, the proof is evident in the event $X_1,X_2\in\{\mathbf{R},\varnothing\}$, we therefore assume that $X_1,X_2\in\{[n,\infty):n\in\mathbf{Z}^+\}$, consequently $X_1 = [n_1,\infty)$ and $X_2 = [n_2,\infty)$ for some $n_1,n_2\in\mathbf{Z}^+$, then $X_1\cap X_2 = [k,\infty)$ where $k = \max(n_1,n_2)$.
$\blacksquare$