Assuming $n > m$ we have

$$\begin{aligned} |n-m| \geqslant \sqrt{n}+\sqrt{m} & \Leftrightarrow (\sqrt{n}-\sqrt{m})(\sqrt{n}+\sqrt{m}) \geqslant \sqrt{n}+\sqrt{m} \\ & \Leftrightarrow \sqrt{n} \geqslant \sqrt{m}+1 . \end{aligned}$$

Let $A_{k} = \left\{ k^2, k^2 + 1, \ldots, (k+1)^2 - 1 \right\}$. Note that each $A_{k}$ can contain at most two elements since if $n, m \in A_{k}$ with $n > m$ then

$$\sqrt{n} - \sqrt{m} \leqslant \sqrt{(k+1)^2 - 1} - \sqrt{k^2} < (k+1) - k = 1$$

In particular, since $S \subseteq A_{1} \cup \cdots \cup A_{44}$, we have $|S| \leqslant 2 \cdot 44 = 88$.

On the other hand, we claim that $A = \left\{ m^2 : 1 \leqslant m \leqslant 44 \right\} \cup \left\{ m^2 + m : 1 \leqslant m \leqslant 44 \right\}$ satisfies the properties and has $|A| = 88$. We check property (b) as everything else is trivial.

So let $r, s, t$ be three elements of $A$ and assume $r < s < t$. There are two cases for $r$.

(i) If we have that $r = m^2$, then $t \geqslant (m+1)^2$ and so $\sqrt{t} - \sqrt{r} \geq 1$ verifying (b).

(ii) If we have that $r = m^2 + m$, then $t \geqslant (m+1)^2 + (m+1)$ and

$$\begin{aligned} \sqrt{t} \geqslant \sqrt{r} + 1 & \Leftrightarrow \sqrt{(m+1)^2 + (m+1)} \geqslant \sqrt{m^2 + m} + 1 \\ & \Leftrightarrow m^2 + 3m + 2 \geqslant m^2 + m + 1 + 2\sqrt{m^2 + m} \\ & \Leftrightarrow 2m + 1 \geqslant 2\sqrt{m^2 + m} \\ & \Leftrightarrow 4m^2 + 4m + 1 \geqslant 4m^2 + 4m . \end{aligned}$$

So property (b) holds in this case as well.