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 $x\in\overline B$.
Let $x\in \bar A$. Then $x$ is a limit point of $A$. Then $\exists$ a sequence $(x\_n)$ in $A$ with $x\_n\to x$. But $(x\_n)$ is also a sequence in $B$ ($x\_n\in A\subset B \forall n$). Thus $x\in \bar B$.