Find the limit points of the set $\{ \frac{1}{n} +\frac{1}{m} \mid n , m = 1,2,3,\dots \}$

Zero is the limit of $\frac1n+\frac1n$ as $n\rightarrow\infty$. And $\frac1n$ is the limit of $\frac1n+\frac1m$ as $m\rightarrow\infty$. Thus $A=\{0\}\cup\{\frac1n\mid n\in\mathbb N\}$ are all limit points. We will show these are the only limit points.

Let $x\not\in A$. Then there is an $\epsilon>0$ such that the interval $[x-\epsilon,x+\epsilon]$ has no point of the form $\frac1n$ (or 0).

Therefore, the sum $\frac1n+\frac1m$, with least one of $\frac1n$ or $\frac1m$ less than $\epsilon/2$, must be at a distance of at least $\epsilon/2$ from $x$. So the only such numbers in the interval $[x-\epsilon/2,x+\epsilon/2]$ must have $\frac1n\geq\epsilon/2$ and $\frac1m\geq\epsilon/2$. So $n\leq2/\epsilon$ and $m\leq2/\epsilon$. Hence there are only a finite number of numbers of the form $\frac1n+\frac1m$ in the interval $[x-\epsilon/2,x+\epsilon/2]$. Thus $x$ is not a limit point.

That's right. For the set: $$ S=\{\frac1n+\frac1m\mid n,m\in \mathbb{N}\} $$ the set $A$ below is the set of all limit points. $$ A=\{0\}\cup\{\frac1a \mid a\in\mathbb{N}\} $$ I came up with a proof inspired by that of Gregory Grant. In fact mine is another way to put his. To show that $A$ contains every limit point of $S$, let's assume real number $x$ and $x\notin A$. We have to show that $x$ is not a limit point for $S$.

Since $x\notin A$, there exists an $\epsilon>0$ for which the interval $I=(x-\epsilon,x+\epsilon)$ has no real number of the form $\frac1a$ where $a\in \mathbb{N}$. Also, $\epsilon$ is such that $0\notin I$, and hence, $x-\epsilon>0$. These are hold for the interval $I'=(x-\frac{\epsilon}{2},x+\frac{\epsilon}{2})$, too. Our goal is to show that there are only finite number of elements both in $S$ and $I'$, if any. To do so, assume $y=\frac1m+\frac1n$ ($n,m\in \mathbb{N}$) to be an element of both $S$ and $I'$, and without loss of generality, $\frac1n\leq\frac1m$. Since $I$ has no real number of the form $\frac1a$ ($a\in \mathbb{N}$), then: $$\frac1n\leq\frac1m\leq x-\epsilon\label{eq1}\tag{1}$$ For $y\in I'$ it is necessary that: $$x-\frac{\epsilon}2<\frac1n+\frac1m\label{eq2}\tag{2}$$ and since $\frac1n\leq\frac1m$, we have $x-\frac{\epsilon}2<\frac2m$, or: $$0<\frac{1}2(x-\frac{\epsilon}2)<\frac1m$$ So there are only finite natural numbers for $m$. On the other hand, using (\ref{eq1}) and (\ref{eq2}): $$x-\frac{\epsilon}2<\frac1n+x-\epsilon$$ or: $$0<\frac{\epsilon}2<\frac1n$$ which shows there are finite natural numbers for $n$. Thus, there are only finite elements like $y$ which belong to both $S$ and $I'$, and hence, $x$ cannot be a limit point for $S$.

Yes, this is true.

To show that each of this is a limit point, you exhibit a sequence in your set that converges to it.

To show that there are no others, you might start by noting that for any $\epsilon > 0$, there are only finitely many points where both $1/m$ and $1/n$ are greater than $\epsilon$.

To show that each limit point of $S$ is in $A,$ we may proceed as follows. Let $x$ be a limit point of $S.$ There is a sequence $(x_n)$ in $S\setminus \{x\}$ such that $x_n\to x.$ Now, each term $x_n=1/m+1/k$ for some $m,k\in\mathbb{N}.$ We look at the sequences $(1/m)$ and $(1/k).$ If both these sequences are eventually constant, then so is $x_n;$ in that case, a tail of this sequence consists of $x$ only. This is not possible since $x_n\in S\setminus\{x\}.$ If exactly one of the sequences is eventually constant, then the other sequence, which is in the form $(1/n)$ converges to $0$ so that the constant is $x$ itself. Now, $x$ is in the form $1/m$ for some $m.$ If none of the sequences $(1/m)$ and $(1/k)$ is eventually constant, then both of them converge to $0;$ so that $x_n\to 0.$ Hence $x\in A.$