Hope I do this correctly

Abbreviating $|S|=n$, it suffices to prove the statement when $n \geq 10^{100}$ by making $c \leq 10^{-100}$. Scale so that $d_{\text{min}}=1$. For a non-intersecting polygon $\mathcal{P}$ let $[\mathcal{P}]$, $P_\mathcal{P}$, and $s(\mathcal{P})$ denote the area, perimeter, and number of sides of $\mathcal{P}$ respectively. Then, it is easy to prove that for a convex polygon $\mathcal{P}$, we can only place at most $$\frac{1}{\pi}([\mathcal{P}]+P_{\mathcal{P}}+(s(\mathcal{P})-2)\pi)$$nonoverlapping circles of radius $1$ whose centers lie in the interior of $\mathcal{P}$. Consider only the half of the plane, demarcated by the extension of the maximal-length segment $\overline{AB}$ to a line, that contains at least $0.5n$ points, and rotate so that $\overline{AB}$ is horizontal, with $A$ to the left of $B$. Let the convex hull of the points on this half-plane be $\mathcal{C}=P_1\ldots P_k$ in that order with $P_1=A,P_k=B$. Then, $P_{\mathcal{C}}$ is at most the perimeter of the region bounded by $\overline{AB}$ and the two circles centered at $A,B$ with radius $AB$, so $P_1P_2+\cdots+P_{k-1}P_k \leq \frac{2\pi}{3}AB \leq \frac{2\pi}{3}cn$; as a corollary, since $P_iP_{i+1}\geq 1$ for each $i$, we get $k \leq \frac{2\pi}{3}cn$. Now, let $t$ be the largest integer such that $\overline{P_{t-1}P_t}$ has positive slope. It is then not hard to prove that for all $i<t$ we have $P_1P_i<P_1P_{i+1}$, while for all $i \geq t$ we have $P_iP_k>P_kP_{i+1}$. We now construct, in order, polygons $\mathcal{P}_1,\ldots,\mathcal{P}_p$, each of which is the convex hull of some $P_i,\ldots,P_j$, as follows. Let $c_1>0$ be a constant which we will determine later. We initialize a counter $u$ to $1$. Then, at each time step, we let $v\geq u$ be maximal such that $P_uP_v \leq \sqrt{c_1n}$, or let $v=t$ if either $v>t$ or no such $v$ exists. Then, we append the convex hull of $P_u,\ldots,P_v$ to $\mathcal{P}_1,\ldots$ and set $u:=v$, unless $v \leq u+1$ in which case we don't append anything and update $u:=u+1$. Likewise, we also construct polygons $\mathcal{Q}_1,\ldots,\mathcal{Q}_q$ with the same value of $c_1$, but we instead initialize a $u$ to $k$ and decrease it at each step until it equals $t+1$. Let $m$ be arbitrary, and say $\mathcal{P}_m$ is the convex hull of $P_i,\ldots,P_j$ with $i<j$. Then if $\theta_m$ is the measure of $\angle P_{i+1}P_iP_j$ in radians, the entirety of $\mathcal{P}_m$ is contained within the sector of a circle with radius $\sqrt{c_1n}$ bounded by $\overline{P_iP_{i+1}}$ and $\overline{P_iP_j}$, which has area $\pi c_1n\theta_m$, so $[\mathcal{P}_m] \leq c_1\theta n\theta_m$. Furthermore, through "arguments of lines" it's easy to see that $\theta_1+\cdots+\theta_p<\pi/2$, so $[\mathcal{P}_1]+\cdots+[\mathcal{P}_p] \leq c_1\pi^2n/2$. The analogous result is true for the $\mathcal{Q}_i$, so all in all we have $$[\mathcal{P}_1]+\cdots+[\mathcal{P}_p]+[\mathcal{Q}_1]+\cdots+[\mathcal{Q}_q] \leq \pi^2c_1n.$$Furthermore, for each $\mathcal{P}_m=P_i\ldots P_j$ we have $P_{\mathcal{P}_m} \leq 2(P_iP_{i+1}+\cdots+P_{j-1}P_j)$. Then, $$P_{\mathcal{P}_1}+\cdots+P_{\mathcal{P}_p}+P_{\mathcal{Q}_1}+\cdots+P_{\mathcal{Q}_q} \leq 2(P_1P_2+\cdots+P_{k-1}P_k) \leq \frac{4\pi}{3}cn.$$Finally, we have $s(\mathcal{P}_m)-2=j-i-1 \leq j-i$, so $$(s(\mathcal{P}_1)-2)+\cdots+(s(\mathcal{P}_p)-2)+(s(\mathcal{Q}_1)-2)+\cdots+(s(\mathcal{Q}_q)-2) \leq k \leq \frac{2\pi}{3}cn.$$Thus, the number of elements of $S$ in $\mathcal{D}':=\mathcal{P}_1 \cup \cdots \cup \mathcal{P}_p \cup \mathcal{Q}_1 \cup \cdots \cup \mathcal{Q}_q$ is at most $\left(\pi c_1+\left(\frac{4}{3}+\frac{2\pi}{3}\right)c\right)n$, while elements of $S$ lying in $\mathcal{D}:=\mathcal{C} \setminus \mathcal{D}'$ is at least $n/2$ minus this. We now upper bound the number of edges of $\mathcal{D}$. Any pair of adjacent edges that contains neither $P_1P_k$ or $P_tP_{t+1}$ (note that both of these are actually edges of $\mathcal{D}$) must have lengths summing to at least $\sqrt{c_1n}$ by construction. Since the sum of all edge lengths is at most $\left(\frac{2\pi}{3}+1\right)cn$, it follows that $\mathcal{D}$ has at most $$\left(\frac{4\pi}{3}+2\right)\frac{c}{\sqrt{c_1}}\sqrt{n}+4$$edges: otherwise, we can find $\left(\frac{2\pi}{3}+1\right)\frac{c}{\sqrt{c_1}}\sqrt{n}$ "good" disjoint pairs of adjacent edges and sum their lengths to get a contradiction. Then, by Pigeonhole on any triangulation of $\mathcal{D}$, we find that some triangle contains at least $$\frac{\left(\frac{1}{2}-\pi c_1-\left(\frac{4}{3}+\frac{2\pi}{3}\right)c\right)n}{\left(\frac{4\pi}{3}+2\right)\frac{c}{\sqrt{c_1}}\sqrt{n}+4}$$points in it. If we choose $c \ll c_1 \ll 1$ so that $c/\sqrt{c_1} \ll 1$, this above quantity is at least $\sqrt{n}$ for $n \geq 10^{100}$. $\blacksquare$