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: 
Suppose $A \subseteq B$.

Then always $B \subseteq \overline{B}$, and $\overline{B}$ is closed. Both are true from the definition of $\overline{B}$. 

Now, $\overline{A} = \bigcap \{C: C \text{ is closed and } A \subseteq C\}$, and
as we saw that $\overline{B}$ is closed *and* contains $A$ (as $A \subseteq B \subseteq \overline{B}$) the set $\overline{B}$ is just *one* of the sets we take the intersection of in this definition of $\overline{A}$ and as the intersection of a (non-empty) family is a subset of each set we are intersecting, by definition, we thus have $\overline{A} \subseteq \overline{B}$.
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 $x\in \bar B$.