limit points of $\{m+\frac{1}{n}:m\in\mathbb{N},n\in\mathbb{N}\}$ 
Let $S=\{m+\frac{1}{n}:m\in\mathbb{N},n\in\mathbb{N}\}$. Find the derived set, $S'$ of $S$.

My approach was simple; for each $m\in\mathbb{N}$, I define a set $A_m=\{m+\frac{1}{n}:n\in\mathbb{N}\}$. Then, there are countable number of $A_m$'s.
We can write $S=A_1\cup A_2\cup...A_m\cup...(m\in\mathbb{N})$. $S$ is a union of countable number of sets therefore, $S'=A_1'\cup A_2'\cup...A_m'\cup...(m\in\mathbb{N})$.
For each $m\in\mathbb{N}, A_m'=\{m\}$.
$\therefore S'=\{m:m\in\mathbb{N}\}=\mathbb{N}$

Is there a better way to prove this? Besides proving that $A_m'=\{m\}$, do I need to prove any other statement/assumption I made?

EDIT: As pointed out by @DonThousand in the comment section, $S'=A_1'\cup A_2'...$ is true for finite union and may/not be true for countable union of  sets so, a correct proof is also asked for.
 A: Firstly, the claim that limit points of a countable union is the union of the limit points of a countable collection is false.  Note that if we define $A_m=\{\frac nm\mid n\in\mathbb Z\}$, then clearly $\bigcup\limits_{m\in\mathbb N}A_m=\mathbb Q$. So, $$\big(\bigcup\limits_{m\in\mathbb N}A_m\big)'=\mathbb R$$ But since each $A_i$ is discrete, $A_i'=\varnothing$ for all $i$. So, $$\bigcup\limits_{m\in\mathbb N}A_m'=\varnothing$$
So how do we actually solve this? In two parts:
Part 1: $\mathbb N\subset S'$
Let $m\in\mathbb N$. Let $\epsilon>0$. By the Archimedian property, there exists $n\in\mathbb N$ such that $\frac1n<\epsilon$. So, $m+\frac1n\in(m-\epsilon,m+\epsilon)$.
Part 2: $S'\subset\mathbb N$
Let $r\in\mathbb R^+$ such that $r\notin\mathbb N$. $r=m+c$, where $m\in\mathbb N\cup\{0\}$, $c\in(0,1)$.
By the Archimedian principle, there exists $x\in\mathbb N$ such that $\frac1x<c$. Consider the set of all $x\in\mathbb N$ such that $\frac1x<c$. As $\mathbb N$ is well ordered, this set has a lower bound. Call this natural $n$. Hence, we know that $\frac1n<c<\frac1{n-1}$ (clearly $n>1$). Consider $\epsilon=\min(\frac1{n-1}-c,c-\frac1n)$. By definition, $\frac1n,\frac1{n-1}\notin B(c,\epsilon)=(c-\epsilon,c+\epsilon)$.
We know that all fractions of the form $\frac1x$, $x\in\mathbb N$ have the property that $\frac1x\leq\frac1n$ or $\frac1x\geq\frac1{n-1}$, so no fraction of this form is in $B(c,\epsilon)$. So, $c$ is not a limit point of the set, so neither is $r=m+c$. 
A: Some general theory can be applied to this problem.

Let there be given a family $(H_i)_{\,i \in \Bbb N}$ of closed sets of $\Bbb R$ such that for each $i \in \Bbb N$,
$\tag 1 H_i \subset [i,i+1]$
Proposition 1: The countable union $H = \cup \,H_i$ is a closed set.
Proof
Let $h$ be any limit point of $H$.
Since $(-\infty,1)$ is an open interval disjoint from $H$, we must have $h \ge 1$. 
If $h$ is not an integer then there exist an integer $m \ge 1$ such that the point $x$ belongs to the open interval $(m,m+1)$. This open interval is disjoint from every set $H_i$ when $i \ne m$. It follows that $h$ is a limit point of $H_i$. Since $H_i$ is closed, $h$ would have to belong to $H_i$.
If $h = 1$, the open interval $(.5,1.5)$ contains $h$ and is disjoint form every $H_i$ with $i \ge 2$.
So the same argument shows that $h \in H_1$.
If $h$ is an integer greater than $1$, then, in a similar fashion one must have that $h \in H_{h-1}$ or $h \in H_h$.
We've accounted for all limit points of $H$ and conclude that $H$ is closed. $\quad \blacksquare$
