Prove that the set $\Big\{ 1/(n+1): n \in \mathbb{N} \Big\} \cup \big\{ 0 \big\} $ is closed.

Let $A = \{ \frac{1}{n+1}: n\in\mathbb{N} \} \cup \{0\}$. Then $$A^{c} = \left(\bigcup_{n=1}^{\infty}\left(\frac{1}{n+1},\frac{1}{n}\right)\right) \cup (-\infty,0) \cup (1,\infty)$$ is a countable union of open intervals (which are open sets), hence it is open. Therefore $A$ is closed.

$\textbf{Edit}$: my answer assumed that $\mathbb{N}$ includes $0$. If your convention for $\mathbb{N}$ does not include $0$, then we would have $$A^{c} = \left(\bigcup_{n=2}^{\infty}\left(\frac{1}{n+1},\frac{1}{n}\right)\right) \cup (-\infty,0) \cup \left(\frac{1}{2},\infty\right)$$ and the conclusion is the same.

Thanks for clarifying that $\mathbb{N}$ does not contain $0$ in your world. Here's your answer. Let's clean it up a bit.

Let $A = \{\frac{1}{n} \mid n \geq 2\} \cup \{0\}$. This is the same set as yours but maybe presented a little clearer (all symbols $n$ are integers). By writing it out, your complement is everything EXCEPT $$ \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, 0. $$ Thus the complement of $A$ is $$ (-\infty, 0) \cup \left(\frac{1}{2}, \infty\right) \cup \bigcup_{k=2}^\infty \left(\frac{1}{k+1}, \frac{1}{k}\right). $$ This is a union of open sets, so it's open. Hence $A$ is closed.

Of my three "terms" the first gets rid of the negatives, the second gets you everything above $\frac{1}{2}$, and the last gets all those teeny intervals between.