Solution:
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 $A_{k}=\{k^{2}, k^{2}+1, \ldots,(k+1)^{2}-1\}$. Note that each $A_{k}$ can contain at most two elements of $A$ 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 $A \subseteq A_{1} \cup \cdots \cup A_{44}$, we have $|A| \leqslant 2 \cdot 44=88$.

On the other hand we claim that $A=\{m^{2}: 1 \leqslant m \leqslant 44\} \cup\{m^{2}+m: 1 \leqslant m \leqslant 44\}$ 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
$$\begin{aligned} \sqrt{t} \geqslant \sqrt{r}+1 & \Leftrightarrow \sqrt{(m+1)^{2}+(m+1)} \geqslant \sqrt{m^{2}+m}+1 \\ & \Leftrightarrow m^{2}+3 m+2 \geqslant m^{2}+m+1+2 \sqrt{m^{2}+m} \\ & \Leftrightarrow 2 m+1 \geqslant 2 \sqrt{m^{2}+m} \\ & \Leftrightarrow 4 m^{2}+4 m+1 \geqslant 4 m^{2}+4 m . \end{aligned}$$
So property (b) holds in this case as well.