$\newcommand{\cl}{\operatorname{cl}}$There are simpler examples than the one that I’m about to give, but the ones that I can think of at the moment all require some mathematical background that you may not have.
Say that a set $A\subseteq\Bbb Z^+$ is closed under addition if $a+b\in A$ whenever $a,b\in A$. For $A\subseteq\Bbb N$ define the closure of $A$ to be
$\cl A=\bigcap\{C\subseteq\Bbb Z^+:A\subseteq C\text{ and }C\text{ is closed under addition}\}\;,$
the intersection of all supersets of $A$ that are closed under addition.
Now let $A=\{n\in\Bbb Z^+:n\text{ is even}\}$, and show that $A=\cl A$.
Let $\mathscr{C}=\{C\subseteq\Bbb Z^+:A\subseteq C\text{ and }C\text{ is closed under addition}\}$. Clearly $A\in\mathscr{C}$, so $\cl A=\bigcap\mathscr{C}\subseteq A$. On the other hand, $A\subseteq C$ for all $C\in\mathscr{C}$, so $A\subseteq\bigcap\mathscr{C}=\cl A$. Putting the pieces together, we have $A=\cl A$.
I think that you’ll find it hard to come up with a proof that uses a chain of double implications.
Or show that $\cl\{2\}=A$.
Let $\mathscr{C}=\{C\subseteq\Bbb Z^+:2\in C\text{ and }C\text{ is closed under addition}\}$, so that $\cl\{2\}=\bigcap\mathscr{C}$. Clearly $A\in\mathscr{C}$, so $\cl\{2\}\subseteq A$. To show that $A\subseteq\cl\{2\}$, suppose not. Then $A\setminus\cl\{2\}\ne\varnothing$, so let $m=\min(A\setminus\cl\{2\})$. Clearly $m\ne 2$, since $2\in\{2\}\subseteq\cl\{2\}$, so $m\ge 4$, and $m-2\in A$. Moreover, $m$ is the smallest member of $A$ that is not in $\cl\{2\}$, so $m-2\in\cl\{2\}$, and therefore $m-2\in C$ for every $C\in\mathscr{C}$. But we also have $2\in C$ for every $C\in\mathscr{C}$, and each $C\in\mathscr{C}$ is closed under addition, so $2+(m-2)=m\in C$ for every $C\in\mathscr{C}$. But then $m\in\bigcap\mathscr{C}=\cl\{2\}$, contradicting the choice of $m$. This contradiction shows that $A\setminus\cl\{2\}$ must actually be empty, i.e., that $A\subseteq\cl\{2\}$. Putting the pieces together, we have $A=\cl\{2\}$, as desired.
Here it’s even harder to come up with a proof that uses a chain of double implications.