About $\mathring{\overbrace{[-4, 0]\cup\left\{ \frac{1}{n}: n \geq1 \right\} }}$

Note that $(-4,0)\subset S$ and it is an open set. Therefore, $(-4,0)\subset\mathring S$.

Now, let $x\in S\setminus(-4,0)$. Then $x=0$ or $x=\frac1n$, for some $n\in\Bbb N$. But $0\notin\mathring S$, because any neighborhood $V$ of $0$ contains some interval $(0,\varepsilon)$ and this interval has numbers $y$ which are not of the form $\frac1n$, for some $n\in\Bbb N$. But then $y\in V\setminus S$, and therefore $V\not\subset S$. So, $0\notin\mathring S$. And if $x=\frac1n$ for some $n\in\Bbb N$, then take $W=\left(\frac1{n+1},\frac1{n-1}\right)$ (if $n>1$) or $V=\left(\frac12,\infty\right)$ (if $n=1$). Then $W$ is a neighborhood of $\frac1n$, but $\frac1n$ is the only element of $S\cap W$. So, if $V$ is any neighborhood of $\frac1n$, then $W\cap V$ is a neighborhood of $\frac1n$ which is not a subset of $S$. So, $\frac1n\notin\mathring S$.

So, $(-4,0)=\mathring S$.