Let $S_K^n$ be the stereographically projected sphere (on the Euclidean plane $E^n$). Where $n$ is the dimensionality of the sphere and $K$ is its sectional cruvature. Then the distance function for $x,y\in S_K^n$ is: $$ d_K(x,y)= \frac{1}{\sqrt{K}} \arccos\left( \frac{ 4\langle x,y\rangle + (||Kx||_2^2-1)(||Ky||_2^2-1) }{ (||Kx||_2^2+1)(||Ky||_2^2+1) } \right) $$ Prove or disprove that in the limit we have: $$ \lim_{K\to 0}d_K(x,y) =c||x-y||_2, \tag{*} $$ where $c$ is a constant factor (most likely $2$).
This would mean that in the limit we recover Euclidean-like geometry.
Note that the distance function above results from the stereographic projection of the sphere through the north pole: $S_K^n\setminus\{\text{north pole}\}\to E^n$.
I'm not sure that this can be shown. However, for the Poincaré disk (if $d_K$ was the distance on the Poincaré disk and $x,y$ would be on the disk) one can show that the upper relationship $(*)$ holds with $c=2$.
One might have to add that $x$ and $y$ lie in the southern hemisphere for such a relationship to hold.
The formula you have written for $d_K(x,y)$ does not posses a finite limit as $K\to 0$ for most choices of $x$ and $y$. Indeed, $$ \cos\Bigl[\sqrt{K}\cdot d_K(x,y)\Bigr]=\frac{ 4\langle x,y\rangle + (||Kx||_2^2-1)(||Ky||_2^2-1) }{ (||Kx||_2^2+1)(||Ky||_2^2+1) } . $$ Since $\cos$ is continuous and $$ \lim_{K\to 0}\frac{ 4\langle x,y\rangle + (||Kx||_2^2-1)(||Ky||_2^2-1) }{ (||Kx||_2^2+1)(||Ky||_2^2+1) }=1+4\langle x,y\rangle, $$ it follows that $\sqrt{K}\cdot d_K(x,y)$ converges to a non-zero constant as $K\to 0$, for suitable $x,y$. But since $\frac{1}{\sqrt{K}}\to\infty$ as $K\to 0$, this means that $$d_k(x,y)=\frac{1}{\sqrt{K}}\cdot \sqrt{K}\cdot d_K(x,y)$$ also tends to $\infty$ as $K\to 0$.
Thus, there does not exist any finite $c$ with the property that $\lim_{K\to 0}d_K(x,y) =c||x-y||_2,$ since the left side is infinite for most $x,y$ whereas the right side is always finite.