HINT: Let

$$D=\left\{\sqrt{n+1}-\sqrt{n}:d\in\Bbb Z^+\right\}=\left\{\frac1{\sqrt{n+1}+\sqrt{n}}:d\in\Bbb Z^+\right\}\;.$$

Let $(a,b)$ be any non-empty open interval in $\Bbb R$. If $a<0<b$, then $0\in(a,b)\cap S$, so suppose that $0\le a<b$. Let $\delta=b-a$; there is a $d\in D$ such that $d<\delta$; why?

For each $k\in\Bbb Z^+$ we have

$$k\left(\sqrt{n+1}-\sqrt{n}\right)=\sqrt{k^2(n+1)}-\sqrt{k^2n}\in S\;,$$

so $\{kd:k\in\Bbb Z^+\}\subseteq S$; show that $(a,b)\cap\{kd:k\in\Bbb Z^+\}\ne\varnothing$, thereby showing that $S$ is dense in $\Bbb R^+$, the positive reals.

Once you know that $S$ is dense in $\Bbb R^+$, showing that it’s dense in the negative reals is very straightforward.