Q:

Show that the set $A$ is not an open nor a closed subset of $\mathbb{R}$

I want to show that the set $A=\left \{\frac{1}{n}\mid n\in \mathbb{N}, n>0\right \}$ is not an open nor a closed subset of $\mathbb{R}$. I want to show also that the set $B=A\cup \{0\}$ is not open but closed in $\mathbb{R}$. 
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I have done the following: 
The set $A$ is not open, since $0$ is a limit but it is not included in the set. 
The set $A$ is not closed: For each $\epsilon >0$ there is a $n\in \mathbb{N}$ such that $\epsilon>\frac{1}{n}$. So for each $\epsilon>0$ the intersection $(0-\epsilon, 0+\epsilon)\cap A$ is non-empty. How do we continue? 
If we consider my justification that $A$ is not open shouldn't $B$ be open? 

A:

$A$ is closed $iff$ it contains all its limit points. $0$ is a limit point, not in $A$, therefore $A$ is not closed.
$A$ is not open since, if you take $1 \in A$, then you can't fit a ball of any radius $r$ inside $A$, ie the interval $(1-r,1+r)$ is not in $A$.
$B$ is not open for the same reasons as $A$. $B$ contains all its limit points, so it's closed. 

