Another characterization of an open set of real numbers I'm trying to prove that. Let $A$  be a subset of $\mathbb{R}$. Then 


Prove that: A is open if, and only if, $\forall X \subset \mathbb{R}$ we have that $A \cap \overline{X} \subset \overline{A \cap X}$


Here a notation note: I say that, if $X \subset \mathbb{R}$ then $\overline{X}$ is the closure of the set, i.e, for every sequence $(x_n)_{n \in \mathbb{N}}$ such that $x_n \in X$,  $\forall n \in \mathbb{N}$ that converges to a number $r \in \mathbb{R}$ then $r \in \overline{X}$.
 A: The result holds not just for $\Bbb R$, but for all topological spaces, though one cannot use sequences to prove the general result.

Let $Y$ be a space. Then $A\subseteq Y$ is open if and only if $A\cap\operatorname{cl}X\subseteq\operatorname{cl}(A\cap X)$ for each $X\subseteq Y$.

Suppose first that $A\cap\operatorname{cl}X\subseteq\operatorname{cl}(A\cap X)$ for each $X\subseteq Y$. Let $X=Y\setminus A$; then 
$$A\cap\operatorname{cl}X\subseteq\operatorname{cl}(A\cap X)=\varnothing\;,$$
so $A\subseteq Y\setminus\operatorname{cl}X\subseteq Y\setminus X=A$. It follows that $A=Y\setminus\operatorname{cl}X$ and hence that $A$ is open.
Now suppose that $A$ is open, and let $X\subseteq Y$ be arbitrary. Suppose that $x\in A\cap\operatorname{cl}X$, and let $U$ be any open nbhd of $x$. Then $A\cap U$ is an open nbhd of $x$, and $x\in\operatorname{cl}X$, so $(A\cap U)\cap X\ne\varnothing$, and $x\in\operatorname{cl}(A\cap X)$.
A: Suppose $A$ is open. Let $x \in A \cap \overline X$. Hence, there is a sequence $x_n$ in $X$ converging to $x$, furthermore, there is a neighbourhood of $x$, call it $V$, such that $V \subset A$, by openness. Since $V$ is a neighbourhood of $x$, then by definition of convergence, for large enough $N$, $x_N \in V$. 
Now,we have a sequence $x_{n+N}$, which converges to $x$ (subsequence of convergent sequence), and which lies entirely in $V$ (hence $A$) and in $X$. Hence, $x \in \overline{A \cap X}$, since we found a sequence in $A \cap X$ converging to $x$.
For the converse, suppose that for all subsets $X$, $A \cap \bar{X} \subset \overline{A \cap X}$.
Suppose that $x$ is not an interior point of $A$, that is, every neighbourhood of $x$ intersects $A^c$. Let $V_n$ be the ball of radius $\frac 1n$ around   $x$, then since $V_n \cap A^c \neq \emptyset$, there is a point $x_n \in V_n \cap A^c$. Now, $x_n$ converges to $x$ because the $V_n$ are arbitrarily small neighbourhoods. Hence, $x \in \overline{A^c}$.
But then, $A \cap \overline {A^c}$ is non-empty, because $x \in A \cap \overline {A^c}$. However, $\overline {A \cap A^c} = \overline{\emptyset} = \emptyset$. this contradicts our hypothesis with $X = A^c$.
Hence, $x$ is an interior point of $A$. This completes the other direction.
