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 $x\in\overline{A}$ then by definition for every $\varepsilon>0$ the ball $B(x,\varepsilon)\cap A\neq\emptyset$. Since $A\subset B$ then $$\emptyset\neq B(x,\varepsilon)\cap A=B(x,\varepsilon)\cap (A\cap B)=(B(x,\varepsilon)\cap B)\cap A\Rightarrow B(x,\varepsilon)\cap B\neq \emptyset$$
hence $x\in\overline{B}$.

$\textbf{For a general topological space:}$ Let $X$ be a topological space and $A, B\in X$ such that $A\subset B$. If $x\in \overline{A}$ then by definition for each neighborhood $N$ of $x$ we have $N\cap A\neq\emptyset$. We need to show $\overline{A}\subset\overline{B}$. From the argument above since $A\subset B$ then 
$$N\cap A=N\cap (A\cap B)=(N\cap B)\cap A\neq\emptyset$$
implies $N\cap B\neq\emptyset$ for any neighborhood $N$ of $x$. Hence $x\in\overline{B}$.