On $X={\bf R}^2$ or $S^2(1)$, we have a triangle $\triangle ABC$ whose perimeter is small. On $D\in \overline{BC}$, let $$ r_1:=|BD|,\ r_2:=|CD| $$
Consider spheres $S(B,r_1),\ S(C,r_2),\ S(A,r)$. Here by choosing $r$ suitably we have two intersection points $$ X\in S(B,r_1)\cap S(A,r),\ Y\in S(C,r_2)\cap S(A,r)$$ in the interior of $\triangle ABC$. Then show that $ \overline{BX},\ \overline{CY}$ meet in the interior of $\triangle AXY $. I can not convince that this is true.
(1) Particularly note that if $Y\in \overline{AC}$, then $\overline{BX}$ pass through some point in $\overline{AY}$. So this case is compatible with this.
(2) So we must show that $\overline{BX}$ does not intersect with the arc $XY$ of $S(A,r)$.
Thank you for your attention.
If $X$ is in the interior of $\triangle ABC$, then $X$ is in the inside of $\angle CBA$. So any point on the ray $AX$ must be on the correct (i.e. inside) side of both $AB$ and $BC$. Likewise any point on $BY$ must be on the correct side of $AC$ and $BC$. Hence they must intersect on the correct side of all three. You might want to rule out the case that the lines intersect on the wrong side of the starting point, not the rays I described above. But a simple angle comparison ($\angle CBX<\angle CBA$ and accordingly at $C$) will verify this.