The container
Begin with the square [0,3] × [0,3] and remove its four 1 × 1 corner squares. The remaining plus sign consists of a central unit square and four arms of width 1. Every disc must lie wholly in this closed region.
Construction: radius 1/2 is attainable
Put five centres at the middle of the plus sign and at the four ends of its arms:
(1.5, 0.5), (0.5, 1.5), (1.5, 1.5), (2.5, 1.5), (1.5, 2.5).
Radius-1/2 discs at these centres lie in the plus sign. Every pair of centres is at distance at least 1, so their interiors do not overlap. Taking any two, three or four of the five discs gives the corresponding smaller construction. Hence rₙ ≥ 1/2 for 2 ≤ n ≤ 5.
Upper bound, step 1: a larger centre cannot enter an arm
Assume r > 1/2. If a centre (x,y) lies in the bottom arm, then 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1. Its distances from the two removed bottom corner squares are x−1 and 2−x. Avoiding both corners therefore requires
x − 1 ≥ r, 2 − x ≥ r ⇒ 1 ≥ 2r,
which contradicts r > 1/2. The other three arms are identical by symmetry. Thus every feasible centre for a disc of radius greater than 1/2 lies in the central unit square [1,2]².
Upper bound, step 2: the feasible centre set is too small
Translate the central square so that its centre is the origin; it becomes [−1/2,1/2]². Let p=(u,v) be a feasible centre and put a=|u|, b=|v|. Choose the corner c of this square in the same quadrant as p. That corner touches a removed square, so the disc must stay at least r away from it:
(1/2 − a)² + (1/2 − b)² ≥ r².
Suppose, for a contradiction, that p is also at least r from the origin. Then a²+b² ≥ r². Adding the two inequalities gives
a² + (1/2 − a)² + b² + (1/2 − b)² ≥ 2r² > 1/2.
But 0 ≤ a,b ≤ 1/2, and for every t in that interval, t²+(1/2−t)² ≤ 1/4. The left-hand side is therefore at most 1/2, a contradiction. Consequently every feasible centre satisfies ‖p‖ < r.
For any two feasible centres p and q, the triangle inequality now yields
‖p − q‖ ≤ ‖p‖ + ‖q‖ < 2r.
Two radius-r discs with centres closer than 2r overlap. Therefore, when r > 1/2, the plus sign cannot contain even two such discs. This proves rₙ ≤ 1/2 for every n ≥ 2. Together with the construction, rₙ = 1/2 for 2 ≤ n ≤ 5. ∎