Question: Let $A,\ B$ denote subsets of a space $X$. Prove that if $A\subset B$, then $\overline{A}\subset\overline{B}$.

**My attempt:**

Let $x\in\overline{A}$

$\implies x\in\bigcap\limits\_{\alpha\in K}U\_\alpha$ where $U\_\alpha$ is a closed set of $X$ and $U\_\alpha\supset A$

$\implies x\in U\_\alpha\ \forall\alpha\in K$

How do I proceed?

Answer: 
If $x$ belongs to every closed set containing $A$, then, in particular, $x$ belongs to every closed set containg $B$ (since $A\subset B$) and therefore $x\in\overline B$.
Correct me if wrong.

$A \subset B.$

$\overline{A}= A\cup A'$ , where $A'$ is the set of limit points of $A.$

Need to show that $A' \subset B'$.

Let $x \in A'$. Then every open neighbourhood $K\_r(x)$ contains points $y \in A,$ and $y \not = x.$

But $A \subset B$ :

Thus every open neighbourhood $K\_r(x)$ , $x \in A'$, contains points $y \in B$, and $y \not = x$, i.e. $x \in B'$.

Hence: $A' \subset B'.$

Finally: 

$\overline{A} =A \cup A' \subset B\cup B'= \overline{B}$.