It is easier than it looks. I guess you know that, for any $A$, it is $A\subseteq \text{cl }A$

Thus, we have that

$A\subseteq \text{cl }A$
$B\subseteq \text{cl }B$

and since $A\subseteq B$, it must be

$A\subseteq \text{cl }B$

Now, suppose that, for the sake of a contradiction, it is $\text{cl }B\subset \text{cl }A$

Then, from our last reasoning, we then have

$A\subseteq B\subseteq \text{cl }B\subset \text{cl }A$

But now, $\text{cl }A$ would not be the smallest closed set containing $A$; since $\text{cl }B$ is closed, for it is the intersection of closed sets, and $$\text{cl }B\subse…