I think you misread the meaning of $(\frac1{n+1},\frac1n)$: It is the open interval with endpoints $\frac1{n+1}$ and $\frac1n$.
The set is the union of open sets and hence open, which implies that all points of $X$ are interior. Indeed, if $x\in X$, then $x\in(\frac1{n+1},\frac1n)$ for some $n$, i.e. $\frac1{n+1}<x<\frac1n$. With $\epsilon=\min\{x-\frac1{n+1},\frac1n-x\}>0$ the $\epsilon$ ball around $x$ is in $X$.
The set of cluster point and the closure are both equal to $[0,1]$ as precisely the $x$ with $0\le x\le 1$ have at least one/infintely many other points in every open neighbourhood.
The boundary $\partial X$ is the difference of closure and interior, i.e. $\{\frac1n\mid n\in\mathbb N\}\cup\{0\}$.
Let's call your set $A$.
Isolated point: Recall that $(x,y)$ is an isolated point of $A$ if $(x,y)\in A$ and there is $r>0$ s.t. $(B_r(x,y)\setminus \{(x,y)\})\cap A=\emptyset$, where $B_r(x,y)$ is the open ball centred at $(x,y)$ of radius $r$. Now for any $(x,y)\in A$, and any $r>0$, since $\mathbb{Q}$ is dense in $\mathbb{R}$, then there are $x',y'\in\mathbb{Q}\cap (0,1)\subseteq A$ s.t. $(x',y')\neq(x,y)$ and $|x-x'|<\frac{r}{2}$, and $|y-y'|<\frac{r}{2}$, thus $(x',y')\in B_r(x,y)$. Since $r>0$ is arbitrary, we conclude that $A$ has no isolated point.
Boundary point: $(x,y)$ is a boundary point of $A$ if $B_r(x,y)\cap A\neq\emptyset$ and $B_r(x,y)\cap A^c\neq\emptyset$ for all $r>0$.
Use the similar argument, one can show that the boundary of $A$ is
$$[0,1]\times[0,1]:=\{(x,y)\in\mathbb{R}^2: 0\leq x\leq 1, 0\leq y\leq 1\}$$
Accumulation point $(x,y)$ is an accumulation point of $A$ if for any $r>0$ we have $(B_r(x,y)\setminus \{(x,y)\})\cap A\neq\emptyset$. Again, applying similar argument, the set of accumulation points is exactly $[0,1]\times[0,1]$
Best Answer
HINTS: $\newcommand{\cl}{\operatorname{cl}}$For $n\in\Bbb Z^+$ let $$A_n=\left\{\frac1n+\frac1k:k\in\Bbb Z^+\right\}\;.$$
Clearly $A_n\subseteq A$, so every accumulation point of $A_n$ is an accumulation point of $A$; what is the unique accumulation point of $A_n$?
For $n\in\Bbb Z^+$ let $p_n$ be the unique accumulation point of $A_n$, and let $B=\{p_n:n\in\Bbb Z^+\}$. Every accumulation point of $B$ is an accumulation point of $A$; why? What is the unique accumulation point of $B$?
Show that $\cl B$ is the set of accumulation points of $A$.
In visualizing $A$, you may find it helpful to show that for each $n>1$, $$A_n\setminus\left(\frac1n,\frac1{n-1}\right)$$ is finite (and you can even calculate exactly how many elements it has). In other words, $A_n$ is almost a subset of the interval $\left(\frac1n,\frac1{n-1}\right)$. This makes it a lot easier to see where the accumulation points are.