Let $S$ be a set of $n$ points in the plane with min spacing of 1. Prove $S$ has a subset of $\ge n/7$ points with min spacing of $\sqrt{3}$. I believe I have proven the case $n=8,|T|=2$, but welcome feedback. I need help proving the case for general $|T|>2$.
From the 2003 Canada National Olympiad:

Let $S$ be a set of $n$ points in the plane such that any two points of $S$ are at least 1 unit apart.
  Prove there is a subset $T$ of $S$ with at least $\dfrac{n}{7}$ points such that any two points
  of $T$ are at least $\sqrt{3}$ units apart.


It suffices to prove the claim for $n=8,15,22,\dots\text{ (for which }|T|=2,3,4,\dots)$, as these are the values at which a new point enters $T$.
The $|T| = 2$ case can be proven for $n=8$ (possibly for $n=7$ if you sharpen the half-plane restriction). For any set of points in the plane, there is at least one point on the convex hull of this set. Let this point be $P$. Then all other points must be weakly within (possibly on the boundary) a half-plane through this point. So if we imagine other points in a fan arrangement around $P$, the total fan angle must be $\le 180^\circ$. In the diagram below, there are points $Q_0,Q_1,\dots,Q_6$ and angles $\theta_1,\theta_2,\dots,\theta_6$ ($i$ ranges from $1$ to $6$). 

Assume that it is possible to arrange the seven $Q$ points in a half-plane about $P$, such that each point is within $\sqrt{3}$ of $P$ and at least $1$ from any other points. Points are labeled in clockwise order.
Consider the angle $\theta_i$. By the cosine rule
$\cos\theta_i = \dfrac{|PQ_{i-1}|^2+|PQ_i|^2-|Q_{i-1}Q_i|^2}{2|PQ_{i-1}|\cdot|PQ_i|} = \dfrac{1}{2} \Bigg(\dfrac{|PQ_{i-1}|}{|PQ_i|} + \dfrac{|PQ_i|}{|PQ_{i-1}|}\Bigg) - \dfrac{|Q_{i-1}Q_i|^2}{2|PQ_{i-1}|\cdot|PQ_i|}$. 
The first two terms always sum to at least $1$ (by AM-GM) and the last term can be minimised by taking $|Q_{i-1}Q_i|=1,\:|PQ_{i-1}|=|PQ_i|=\sqrt{3}$. Thus
$\cos\theta_i \ge 1 - \dfrac{1^2}{2\cdot\sqrt3\cdot\sqrt3} = \dfrac{5}{6}$
So $\theta_i \ge \arccos(\frac{5}{6}) \gtrapprox 33.5^\circ$ 
Then $\sum\limits_{i=1}^{6}{\theta_i} \gtrapprox 201^\circ$, so the $Q_i$ cannot all be in the same half-plane and within $\sqrt3$ of $P$, invalidating the original assumption. Thus, there is at least one pair of points at a distance exceeding $\sqrt3$.
 A: Your argument is good until almost the end.  I'm offering a remedy.  

Let $S$ be a given set with the required property.  I shall prove that there exists a subset $T$ of $S$ and $|T|\geq\frac{|S|}{7}$ such that any two distinct points in $T$ are strictly more than $\sqrt{3}$ units apart. 

Let $P$ a point on its convex hull.  As you have said, there exists a closed half plane with $P$ on the boundary such that all other points in $S$ lie within its interior (i.e., not on the boundary).  
First, if there are points $Q_1$, $Q_2$, $\ldots$, $Q_d$ with $d\geq 7$ such that $\left|PQ_i\right| \leq \sqrt{3}$ for all $i=1,2,\ldots, d$, then it is possible to find two points, say $Q_1$ and $Q_2$ such that $0\leq \angle Q_1PQ_2 < \frac{\pi}{d-1}\leq \frac{\pi}{6}$, so $\cos\left(\angle Q_1PQ_2\right) > \cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$.  That is, 
$$
\begin{align}
\left|Q_1Q_2\right|^2
&=\left|PQ_1\right|^2+\left|PQ_2\right|^2-2\left|PQ_1\right|\,\left|PQ_2\right|\,\cos\left(\angle Q_1PQ_2\right) 
\\
&< \left|PQ_1\right|^2+\left|PQ_2\right|^2-2\left|PQ_1\right|\,\left|PQ_2\right|\,\cos\left(\frac{\pi}{6}\right) 
\\
&=\left|PQ_1\right|^2+\left|PQ_2\right|^2-\sqrt{3}\,\left|PQ_1\right|\,\left|PQ_2\right|\,.
\end{align}$$
Note that $1\leq \left|PQ_1\right|\leq \sqrt{3}$ and $1\leq \left|PQ_2\right|\leq\sqrt{3}$.
Due to the convexity of the function $$f(x,y):=x^2+y^2-\sqrt{3}xy$$
for $(x,y)\in[1,\sqrt{3}]\times[1,\sqrt{3}]$, the supremum of $f$  is attained at the extremities of the region $[1,\sqrt{3}]\times[1,\sqrt{3}]$, namely, at 
$$(x,y)=(1,1)\,,\,\,(x,y)=(1,\sqrt{3})\,,\,\,(x,y)=(\sqrt{3},1)\,,\text{ or }(x,y)=(\sqrt{3},\sqrt{3})\,.$$  Hence, the supremum of $f(x,y)$ is $1$, which is attained if $(x,y)=(\sqrt{3},1)$ or $(x,y)=(1,\sqrt{3})$.  That is, $f(x,y)\leq 1$ for all $(x,y)\in[1,\sqrt{3}]\times[1,\sqrt{3}]$.  Consequently, 
$$\left|Q_1Q_2\right|<\sqrt{f\big(\left|PQ_1\right|,\left|PQ_2\right|\big)}\leq 1\,,$$ contradicting the assumption that every pair of points is at least $1$ unit apart.  Therefore, the assumption $d\geq 7$ is false, so $d\leq 6$.
Hence, for every such set $S$, you can find a point $P$, such that, by removing $d\leq 6$ points $Q_1,Q_2,\ldots,Q_d$ around $P$, you get a new configuration where the distance to $P$ from any other points is more than $\sqrt{3}$.  Now, let $S'$ be the new set $S\setminus\left\{P,Q_1,Q_2,\ldots,Q_d\right\}$.  Then, $\left|S'\right|<|S|$, so by induction there is a subset $T'$ of $S'$ such that $\left|T'\right|\geq \frac{\left|S'\right|}{7}$ and that every pair of points in $T'$ are at distance more than $\sqrt{3}$ apart.  Define $T:=T'\cup\{P\}$; then, $|T|\geq \frac{|S|}{7}$ and satisfies the required condition.  The base cases where $|S|\leq7$ are trivial.
