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.
$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Suppose that you can show that $S\setminus\int E=\cl(S\setminus E)$. Then
$$(\cl E)\setminus\int E=\big((\cl E)\cap S\big)\setminus\int E=(\cl E)\cap(S\setminus\int E)=(\cl E)\cap\cl(S\setminus E)\;,$$
which is exactly what you want.
To see where the idea comes from, note that both $(\cl E)\setminus\int E$ and $(\cl E)\cap\cl(S\setminus E)$ are subsets of $\cl E$. If $A$ is any set disjoint from $\cl E$, therefore,
$$(\cl E)\setminus\int E=(\cl E)\cap\cl(S\setminus E)\quad\text{iff}\quad A\cup\Big((\cl E)\setminus\int E\Big)=A\cup\Big((\cl E)\cap\cl(S\setminus E)\Big)\;,$$
and a clever choice of $A$ might give us sets that are easier to think about than $(\cl E)\setminus\int E$ and $(\cl E)\cap\cl(S\setminus E)$. In particular, if we let $A=S\setminus\cl E$, we can compare the set,
$$(S\setminus\cl E)\cup\Big((\cl E)\setminus\int E\Big)\;,$$
which is simply all of $S$ except $\int E$, or $S\setminus\int E$, with the set
$$(S\setminus\cl E)\cup\Big((\cl E)\cap\cl(S\setminus E)\Big)\;,$$
which is simply $\cl(S\setminus E)$, and try to show that these are equal.
Best Answer
Use my definition of boundary in the comment above.
Let $A=(-\infty,\sqrt 2]\cap \mathbb Q$.
Boundary point of $A$ must be a limit points in $A$, therefore $\partial A\subseteq (-\infty,\sqrt 2]$ because $(-\infty,\sqrt 2]$ contains all of the limit points of $A$ ($A$ is contained in $(-\infty,\sqrt 2]$, and $(-\infty,\sqrt 2]$ is closed in $\mathbb R$).
The rationals and irrationals are dense in $(-\infty,\sqrt 2]$, so $(-\infty,\sqrt 2] \subseteq \partial A$.
Therefore $\partial A=(-\infty,\sqrt 2]$.