$6$ points in the interior of a square of side length $2$.

There is no simple way to prove this, because the statement is false!

Place four points at the corners of the square, and two points at $(x,x)$ and $(2-x,2-x)$, where $x\approx 0.48735$ (though any $x\in[0.44,0.5)$ works).

Then the minimum perimeter of any triangle in the configuration is

$$\frac{6+5\sqrt{2}+\sqrt{22-4\sqrt{2}}}{4}\approx4.2784\ldots\approx3\sqrt{2}+0.03578$$

attained by both the triangles consisting of two adjacent corners and the nearest interior point, and the smaller of the two degenerate triangles lying on the diagonal.

(Because this configuration has all triangles with a strictly larger perimeter, we can move any of the points around by epsilon if we wish to e.g. avoid degenerate triangles, or have all points lie in the interior of the square.)