Closure of a $A\subset \mathbb{R}$ in the topology generated by the subbasis 
Let $\tau$ be the topology on $\mathbb{R}$ given by the subbasis consisting of open intervals $(a,\infty)$. Then I have to determine the following:
  
  
*
  
*The closure of any subset of $\mathbb{R}$.
  
*Sequence lemma holds in this topology. That is given any $x\in\bar{A}$, there exists $x_n\in A$ such that $x_n\to x.$
  

Basically this is an example of a non-Hausdorff space in which sequence lemma holds.
My attempt:

Let $x\in\bar{A}$, then given any neighborhood $U_x$ of $x$ $U_x\bigcap A\setminus\{x\}\neq \emptyset.$ Here basis  is same as the subbasis. $U_x=\cup_{a\in\mathbb{R}}(a,\infty).$ We asume that $x\notin A$. Since, 
$$ A\bigcap \left(\cup_{a\in\mathbb{R}}(a,\infty)\right)\neq \emptyset. $$ After that I stuck. 

 A: Starting with some subbase you can find a base by taking finite  intersections of elements of the subbase. Doing so in this case learns us that the collection $\{(a,\infty)\mid a\in\mathbb R\}$ is not only a subbase but is also a base of $\tau$. This simply because it is closed under finite intersections.
Open sets are arbitrary unions of base elements and secondly we learn that: $$\tau=\{(a,\infty)\mid a\in\mathbb R\}\cup\{\varnothing,\mathbb R\}$$
So the collection of closed sets is:$$\{(-\infty,a]\mid a\in\mathbb R\}\cup\{\mathbb R,\varnothing\}$$
That means that the closure of a set $A$ will take the form $(-\infty,\sup A]$ if $\sup A<\infty$ and will be $\mathbb R$ if $\sup A=\infty$. 
If $x\in \bar{A}$ then $x<\sup A$ or $x=\sup A$
If $x<\sup A$  then $a\in A$ exists with $x<a$. 
If $x<a$ then observe then that $a$ will be an element of every $U\in\tau$ that contains $x$.
This because $U=\mathbb R$ or $U=(y,\infty)$  for some $y<x$.
That means that the sequence $(x_n)$ prescribed by $x_n=a$ converges to $x$.
If $x=\sup A$ and $\sup A\in A$ then likewise with $x_n=\sup A\in A$.
If $x=\sup A$ and $\sup A\notin A$ then a sequence $(x_n)$ exists with $x_n\in A$ and $\lim_{n\to\infty}x-x_n=0$.
Then for each $U\in\tau$ with $x\in U$ we have $x_n\in U$ for $n$ large enough, so $\lim_{n\to\infty}x_n=x$.
