Is $M=\{\frac{1}{k}|k \in \mathbb{N}\}$ closed? Is $M=\{\frac{1}{k}|k \in \mathbb{N}\}$ closed?
I think it is closed, but I'm not sure whether my argumentation is correct.
Since $(\frac{1}{k})_{k \in \mathbb{N}}$ is convergent $\Rightarrow$ from Cauchy $ \exists k_0 \in \mathbb{N}: \forall k \geq k_0:|\frac{1}{k}-\frac{1}{k+1}|< \epsilon ,\forall \epsilon \gt 0$
Let $O_k = (\frac{1}{k+1},\frac{1}{k}), \forall k \lt k_0$
and let $O=(-\infty,0) \cup \bigcup_{k=1}^{k_0-1}O_k \cup (1, \infty)$
$\Rightarrow M=\mathbb{R}-O$
Since O is open, M is closed. Is that argumentation correct?

It is not true that $M=\mathbb{R}-O$; the set $\mathbb{R}-O$ still contains all the intervals $(\frac{1}{k+1},\frac{1}{k})$ for $k\geq k_0$, and these intervals are not contained in $M$.
It's possible some of your confusion is coming from the order of the quantifiers in the definition of Cauchy - you have to fix $\varepsilon$ first, and this determines $k_0$. Whatever $\varepsilon$ you choose, your $\mathbb{R}-O$ still differs from $M$ by infinitely many intervals. It doesn't even help that the length of these intervals is bounded by $\varepsilon$ - even if you make sense of taking $\varepsilon$ to $0$, so that "$\mathbb{R}-O$ converges to $M$", this convergence won't preserve the openness of $O$.
As the other answers point out, $M$ has a limit point (i.e. $0$) that it doesn't contain, so it isn't closed.
