19 points on a hexagon

$\alpha=\frac{1}{2}$ is the optimal bound. We just need to prove that for every $\varepsilon>0$ we can take $19$ points in the hexagon such that the distance between any two of them is $\geq\frac{1}{2}-\varepsilon$. Easily done: we place $19$ points in a regular hexagon width side length $1-\delta$ accordingly to the given construction, then apply a homothety bringing the $(1-\delta)$-hexagon into a unit hexagon. Any two "scaled" points inside the unit hexagon will be separated by a distance $\geq \frac{1}{2(1-\delta)}$, so $\alpha\geq\frac{1}{2}$ follows from taking: $$\delta = \frac{2\varepsilon}{1+2\varepsilon}.$$ To prove $\alpha\leq\frac{1}{2}$, we split the original unit hexagon in $18$ congruent cyclic quadrilaterals (boundaries in red): Now, if eight or more points fall in the innermost hexagon, we are ok since we can cover the innermost red hexagon with seven polygons having diameter $\frac{1}{2}$. This gives that we can assume that there are at most seven points in the innermost hexagon, then at least twelve points in the outermost annulus. Since the perimeter of the external boundary of the annulus is six, even in this case there are $2$ points at most $\frac{1}{2}$-apart.

A simple measure-theoretic argument provides a little weaker bound. Assuming that we can place $21$ points inside a unit hexagon in such a way that the distance between any two of them is $\geq\frac{1}{2}$, then we can fit $21$ circles with radius $\frac{1}{4}$, without any overlapping, inside a regular hexagon with side length $1+\frac{1}{4}$. However, that gives a contradiction since: $$ 21\cdot\frac{\pi}{16}>\frac{3\sqrt{3}}{2}\left(1+\frac{1}{4}\right)^2, $$ so in the unit hexagon we can place at most $20$ points in such a way that the distance between any two of them is $\geq\frac{1}{2}$. I hope someone is able to improve this argument to show that to place $20$ circles is also impossible.