EDIT: After talking with a friend about this problem, I realized that I actually just solved the case $n=3$. The general case, however, follows directly: we can take a triangle satisfying the problem's conditions for the length $\frac\alpha n$, and scale it up by a factor of $n$. By extending its long side, we can create $n-1$ rows of points in the triangle's interior. This is more than enough to build a convex lattice $n$-gon, which will satisfy the problem's constraints, being in the interior of a triangle that does.
The problem essentially asks us to prove that, for every set of directions, there exists a convex lattice polygon as “thin” as we want it to in every of those directions. To prove this, we’ll construct a lattice triangle satisfying our constraints, for every choice of $S$, $\alpha$.
To do this, we’ll need two lemmas.
Lemma $\mathbf1$:
For every segment $\ell$ of slope $\frac pq$ connecting two lattice points, with $p$, $q$ coprime integers, there exists a lattice point $A$ at a distance of no more than $\frac 1q$ of $\ell$.
Proof:
Consider the leftmost extreme of $\ell$, with coordinates $(x,y)$. If $\frac pq=\frac01$, then $A=(x+1,y)$ suffices. Otherwise, let $0\leq a<q$ be a multiplicative inverse of $p\bmod q$. The lattice point $A=\left(x+a,y+\frac{ap-1}q\right)$ is at a distance of $\frac 1q$ from the point $\left(x+a,y+\frac{ap}q\right)\in\ell$, so that the distance from $A$ to $\ell$ does not exceed our desired amount. $\square$
Lemma $\mathbf2$:
For every $\bigtriangleup ABC$ and every $m$, if the cevian $AX$ with slope $m$ is contained in $\bigtriangleup ABC$, it is the longest segment with such slope inside of the triangle.
Proof:
Take a segment $PQ$ of slope $m$ inside $\bigtriangleup ABC$. We can extend it to a segment $P’Q’$ whose endpoints are in the perimeter of the triangle. This segment will either be contained in $\bigtriangleup ABX$ or in $\bigtriangleup ACX$. Either way, it will be smaller than $AX$, by Thales. $\square$
We’re now ready to begin our construction.
For every $m\in S$, we can create an associated angle $0\leq m_a<\pi$, that represents the (directed) angle of a line with slope $m$ with respect to the horizontal. We’ll call the set of all of these angles $S_a$.
Now, we can consider an arbitrary $\theta\not\in S_a$. Let $2\varepsilon$ be the smallest distance ($\text{mod }\pi$) from $\theta$ to any element of $S_a$. Since $\mathbb Q$ is dense in $\mathbb R$, we can take a line $\ell$ through two lattice points with slope $\frac pq$, with $p$, $q$ coprime integers, such that $\ell$ makes a (directed) angle with the horizontal between $\theta-\varepsilon$ and $\theta+\varepsilon$ ($\text{mod }\pi$), and such that $$q\geq\frac1{\sin(\varepsilon)\alpha}.$$ By lemma $\mathbf1$, we can take a lattice point $A$ at a distance of $d\leq\frac 1q$ from $\ell$. Then, we can take lattice points $B$ and $C$ in $\ell$ “far enough”, so that the resulting $\bigtriangleup ABC$ contains cevians from $A$ with all slopes in $S$. Finally, by lemma $\mathbf2$, the length of any segment in $\bigtriangleup ABC$ with a slope $m\in S$ will not exceed the length of the corresponding cevian $AX$, which measures $$\frac d{\sin(\angle AXB)}\leq\frac d{\sin(\varepsilon)}\leq\frac1{q\sin(\varepsilon)}\leq\alpha,$$ just as we wanted. $\blacksquare$
Best Answer
I'll assume that you mean the one-sided classifiers which assign $+$ to every point inside the (closed) polygon and $-$ outside.
Note that the proof idea you suggest in your question is a little bit flawed - 1. shattered sets are allowed to be 'special,' since the VC dimension is defined as the largest set of shattered points, and not the largest set of shattered points in general position. For example, in your lower bound, you arranged points on a circle! 2. every pair of points is collinear - you can draw the line joining them! Collinearity is only a non-trivial property for $\ge 3$ distinct points.
For the proof - consider any collection of $2d + 2$ points. There are two cases:
Suppose the points do not lie on a convex $2d + 2$-gon. Then there must be some point $p$ that lies in the convex hull of the remaining points. But then any polygon that contains the rest of the points must also contain $p,$ and so the labelling $p = -,$ everything else = $+$ cannot be achieved.
This leaves the case when the points do lie on a convex $2d+2$-gon. For this, consider the alternating labelling. Travelling around the polygon, we will have $d+1$ instances of the pattern $+ - +$ (in order). But each such pattern requires a line to separate, and the convexity of the arrangement ensures that no line will separate two patterns. Thus, we need at least $d+1$ lines to achieve this labeling - ergo, we cannot do it with $d$ lines.