Show that if $U$ is a nonempty open set in $\mathbb{R}$, then $\overline{U}$ = $\mathbb{R}$. 
Let $\mathbb{R}$ be a topological space with topology $T = \{ U \subseteq \mathbb{R} : U^c \ \text{is finite} \} \bigcup  \{\emptyset \}$. Show that if $U$ is a nonempty open set in $\mathbb{R}$, then $\overline{U} = \mathbb{R}$. 

So I suppose that $U$ is a non open set in $\mathbb{R}$ 
$\Rightarrow$ $U$ is a subset of $\Bbb R$ such that U$^c$ is finite
What is the next step? 
 A: If $U \in T\setminus\{\emptyset\}$, then 
$$
\mathbb{R}\setminus U=\{u_1,\ldots,u_m\}
$$ 
It follows that
$$
U=\mathbb{R}\setminus\{u_1,\ldots,u_m\}.
$$
To prove that $\overline{U}=\mathbb{R}$ it is enough to show that given $x \in \mathbb{R}$, we have $U_x\cap U\ne \emptyset$ for every $U_x \in T$ with $x \in U_x$. So let $U_x \in T$ with $x \in U_x$. Then $U_x=\mathbb{R}\setminus\{x_1,\ldots,x_n\}$. We now have
$$
U_x\cap U=(\mathbb{R}\setminus\{x_1,\ldots,x_m\})\cap(\mathbb{R}\setminus\{u_1,\ldots,u_m\})=\mathbb{R}\setminus(\{x_1,\ldots,x_n\}\cup\{u_1,\ldots,u_m\}).
$$
It is obvious that 
$$
\mathbb{R}\setminus(\{x_1,\ldots,x_n\}\cup\{u_1,\ldots,u_m\})=\mathbb{R}\setminus\{x_1,\ldots,x_n,u_1,\ldots,u_m\}
$$
is non-empty since $\{x_1,\ldots,x_n,u_1,\ldots,u_m\}$ is a finite subset of $\mathbb{R}$.
A: The union of two finite sets is finite. If $U$ and $V$ are non-empty open subsets of $\mathbb{R}$, then can you prove that the complement of $U\cap V$ is finite? Can you now see how to prove that every nonempty open set is dense? Also, can you see where the "nonempty" assumption is used?
I hope this helps!
A: If $U$ is an nonempty open set in $\Bbb R$ then $\Bbb R\smallsetminus U$ finite, say $\{u_1,\ldots,u_r\}$. To show that $\overline U=\Bbb R$, you need to show that every element of $\Bbb R$ is in the closure of $\overline U$, which amounts to showing that for every $x\in\Bbb R$ and every open set $N$ containing $x$, $N\cap U$ is nonempty. But $N$ being open has finite complement, call it $\{n_1,\ldots,n_q\}$. Now, $$\Bbb R\smallsetminus(N\cap U)=\{n_1,\ldots,n_q\}\cup \{u_1,\ldots,u_r\}$$  by De Morgan's laws. The complement, call it $C$, of the above is $N\cap U$. Is this nonempty for any choice of $x$ and $N$? What do you conclude?
ADD Note this works in general: if you endow an infinite set with the cofinite topology, then every non-empty open subset will be dense.
