Determining compactness of the set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$ in metric space $\mathbb R$ Given a set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$. 
My attempt: 
Claim $1:0$ is a limit point of set $S$.
Proof: Given any $\epsilon \gt 0, \exists N_\epsilon \in \mathbb N: \frac1{N_\epsilon}\lt \epsilon/2$. Choosing $m,n \ge N_\epsilon$,we have $\frac1{n}\le \frac1{N_\epsilon}\lt \epsilon/2$ and $\frac1{m}\le \frac1{N_\epsilon}\lt \epsilon/2 $. Thus $0\lt\frac{1}{m}+\frac{1}{n}\lt \epsilon$. Since $\epsilon \gt 0$ is arbitrary, we conclude that $0$ is a limit point of $S$. 
Claim $2$: All elements of set $T=\{1/n: n\in \mathbb N\}$ are also limit points of $S$.
Proof: Given any $\epsilon \gt 0, \exists M_\epsilon \in \mathbb N: \frac1{M_\epsilon}\lt \epsilon\implies \frac1{m}\lt \epsilon \;\;\forall m\ge M_\epsilon\implies -\epsilon\lt \frac1{m}\lt \epsilon$. Hence for any $\frac1{n}\in T$, we have $\frac 1{n}-\epsilon\lt \frac 1{n}+ \frac1{m}\lt \frac1{n}+\epsilon$. Just as previosuly, it follows that all elements of $T$ are limit points of $S$.
Heine-Borel theorem states that a set $S$ in $\mathbb R$ is compact $\iff S$ is closed and bounded. Indeed, in the present case, $S$ is bounded as $2$ is one upper bound. 
Since, $1/n \notin S$ for any $n\in \mathbb N$, $S$ is not closed and therefore, $S$ can't be compact. 
Is my conclusion correct? Thanks.
Edit: How do I show that $S$ does not have limit points except $0$ and $\{1/n: n\in \mathbb N\}$?
 A: As noted in the comments, $T\subseteq S$: for each $\frac1n\in T$ we have $\frac1n=\frac1{2n}+\frac1{2n}$. Even if the definition of $S$ were changed to require $m\ne n$, we could write
$$\frac1n=\frac1{n+1}+\frac1{n(n+1)}$$
for any $n\in\Bbb Z^+$. Thus, if you can show that $0$ and the numbers $\frac1n$ for $n\in\Bbb Z^+$ are the only limit points of $S$, you can conclude that $S$ is compact. I’ll sketch an argument and let you fill in the details.
Let $x\in\Bbb R\setminus S$. If $x<0$, then $(x-1,0)$ is an open interval around $x$ disjoint from $S$, so $x$ is not a limit point of $S$. Similarly, if $x>2$, then $(2,x+1)$ is an open interval around $x$ disjoint from $S$.
Now suppose that $1<x<2$. Since $\frac1n+\frac1m\le 1$ when $n,k\ge 2$,
$$S\cap(1,2)\subseteq\left\{1+\frac1n:n\in\Bbb Z^+\right\}\,,$$
and therefore
$$(1,2)\setminus S=\bigcup_{n\ge 1}\left(1+\frac1{n+1},1+\frac1n\right)\,,$$
so there is a unique $n\in\Bbb Z^+$ such that $\left(1+\frac1{n+1},1+\frac1n\right)$ is an open interval around $x$ disjoint from $S$.
Finally, suppose that $0<x<1$. There is a unique $n\in\Bbb Z^+$ such that
$$\frac1{n+1}<x<\frac1n\,.$$
Suppose that $\frac1k+\frac1\ell\in S\cap\left(\frac1{n+1},\frac1n\right)$, with $k\le\ell$.

*

*Show that $k\le 2n+2$.

*Show that for each positive integer $k\le 2n+2$ there are only finitely many $\ell\in\Bbb Z^+$ such that $\frac1k+\frac1\ell<\frac1n$.

*Conclude that there are $a,b\in S$ such that $a=\max\{s\in S:s<x\}$ and $b=\min\{s\in S:x<s\}$, so that $(a,b)$ is an open interval around $x$ disjoint from $S$.

A: Consider this:
Let $x \not\in S$. We are going to show $x$ can't be a limit point of $S$.  As $S$ is bounded below by $0$ and above by $2$ we don't need so consider $x \le 0$ or $x \ge 2$.
So $x \ne \frac 1m = \frac 1{2m}+\frac 1{2m}$
Now $x > 0$ So there is a minimal $m \in \mathbb N$ so that $\frac 1m < x$.  If $m> 1$ then $\frac 1m < x < \frac 1{m-1}$ and if $m= 1$ then $\frac 1m = 1 < x$
Now $x -\frac 1m > 0$ so there is a minimal $n\in \mathbb N$ so that $\frac 1m + \frac 1n < x$.  And again if $n>1$ then $\frac 1m +\frac 1n < x < \frac 1m + \frac 1{n-1}$ and if $n=1$ then $\frac 1m + \frac 1n = \frac 1m + 1 < x$. (but it's not actually possible for $n = 1$)
It's easy to see that if $m =1$ then $\frac 1m + \frac 1n < x < 2$ so $n > 1=m$, and if $m> 1$ then $\frac 1m +\frac 1n < x < \frac 1{m-1}$ so $\frac 1n < \frac 1{m-1} + \frac 1m = \frac 1{(m-1)m} < \frac 1m$ and $n > m$.  SO $n > m\ge 1$.  So $\frac 1m +\frac 1n < x < \frac 1m +\frac 1{n-1}$
Now pick an $r$ so that $r < \min(x - (\frac 1m +\frac 1n),  (\frac 1m +\frac 1{n-1})- x)$
The $B_r(x)$ can't contain any points of $S$.
So $x$ is not a limit point of $S$ and then only limit points of $S$ (if any) are in $S$ so $S$ is closed.
(I say "if any" because we did nothing to find any limit points and we don't care.  If we actually cared we can see all $\frac 1m$ (including $1$) and $0$ are limit points [as there are infinitely many $\frac 1m + \frac 1n$ is any $B_r(\frac 1m)$] and for any $\frac 1m +\frac 1n\ne \frac 1t; t\in \mathbb N$ we can use a similar argument above is not t limit point.)
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On the other hand a proof by finite subcover could be easier:
If $0 \in U$, an open set, then all but a finite number of $\frac 1t$ are in $U$.  For each of these finite number $\frac 1m \not \in U$ if $\frac 1m \in V_m$, then all but a finite number of $\frac 1m +\frac 1s \not \in V_m$. If each of these finite number of $\frac 1m + \frac 1{n_m}$ is in an open set $W_{m,n_m}$ then we have covered $S$.
For any open cover, $\mathscr U$, of $S$ we can find a $U\in \mathscr U$  so that $0 \in U$.  Then we can have a finite number of $\frac 1m \not \in U$ and we can find a finite number of $V_m \in \mathscr U$ so that $\frac 1m \in V_m$.  And for each of those we can find a finite number of $n_m$ so that $\frac 1m +\frac 1{n_m}\not \in V_m$ we can find a $W_{m,n_m}$ so that $\frac 1m +\frac 1{n_m} \in W_{m,n_m}$.
And $\{U\} \cup \{V_m\}\cup \{W_{m,n_m}\}$ is a finite subcover of $\mathscr U$.
