Two proofs regarding open and closed sets [Question]
1) Let $M=\{\frac{1}{n}:n\in\mathbb{N}\}$. How do I show $M$ is not open in $\mathbb{R}$?
2) Let $K=\{\frac{1}{n}:n\in\mathbb{N}\} \cup \{0\}$. How do i show $K$ is closed in $\mathbb{R}$?
[Solution]
1) 
I've taken the point $1 \in M$ and placed an open ball around it such that $B_r(1)$. 
I want to prove that I can find an $x \in B_r(1)$ such that $x \not\in M$ because then, by definition, $M$ is not open.
I understand that it isnt open, by drawing and looking at the interval. But how do i prove it mathematically?
Could i simply write something like this: 
Let $r>0$ and $x=1+\frac{1}{2}r$ then $x \in B_r(1)$ while $x \not\in M$ therefore $B_r(1) \not\subset M$ and therefore $M$ is not open.
I'm not sure how to argue how/why I defined x as such. Is the above an okay proof and how could I proof/argue how x is defined?
2) To show that $K$ is closed, I will show that $K^c$ is open.
I know $K^c$ is an open union of sets but I'm not certain how to make a proper proof. Here is what I've tried
$K$ will never be negative and will be greater than 1, therefore it will never be within $(-∞;0)$ & $(1;∞)$.
Now I'm not certain what to do.
 A: Here is another proof.
A set $K \subseteq X$ where $X$ is a metric space is closed iff for every sequence $x_n \in K$  such that $x_n \longrightarrow x$ we have that $x \in K$.(It is an easy proposition to prove)
In $A= \{1/n|n \in \mathbb{N} \} \cup \{0\}$ what type of sequences do you have?
Also if you know about compactness,every compact subset of $\mathbb{R}^n$ is closed.
$A$ is compact because ,let $\{A_i|i \in I \}$ be an open cover of $A$.
For some $i_0 \in I$ we have that $0 \in A_{i_0}$ and the sequence $x_n=1/n \in A$ converges to $0$ .We  know that $A_{i_0}$ is open thus there exist $\epsilon >0$ such that $0 \in (-\epsilon,\epsilon) \subseteq A_{i_0}$ 
From the convergence of $x_n$ exist a $N \in \mathbb{N}$ such that $x_N \in (-\epsilon,\epsilon), \forall n \geqslant N$.
For $n<N$ every term of we have that $1 \in A_{s_1}.......\frac{1}{N-1} \in A_{s_{N-1}}$.
Take  the union of these sets with $(-\epsilon,\epsilon)$ and ypou found a finite subcover of $A$.
I hope this proof helps a little. 
It will help if you have encounter compactenness and some theory in metric spaces.
A: This solution is wrong. The second member of the union is not an intersection but an union. In fact we have:
$K^c=(-\infty,0)\cup(\cup_{n=1}^{\infty}(\frac{1}{1+n},\frac{1}{n}))\cup(1,\infty)$.
Since by definition the arbitrary union of open sets is open, we have $K^c$ is open. Therefore $K$ is closed.
