Verification: Closed Set Expands to Fill Space, but Contains No Open Ball $B_\epsilon(0) $?

Question

I have the proof that $C$ closed, convex, symmetric in Banach space $X$ and $\cup_{n \in N \setminus 0} n.C= X $ then $B_\epsilon(0) \subset C$ for some $\epsilon > 0$. I also have the proof for $D$ satisfying the same but without symmetry (just form $C = D \cap (-D)$), so I thought that maybe convexity could be relaxed too, but couldn't prove it.

So then I looked for a counter example and came up with the following:

Consider the set $C$ in the X-Y plane consisting of closed unit disks centered at (0, 1) and (0, -1) together with the closed line segment [(-1, 0), (+1, 0)].

This is closed, being the union of three closed sets and seems to satisfy $\cup_{n \in N \setminus 0} n.C= X $, but any $n.C$ is going to be missing a bit of any open ball $B_\epsilon(0) $.

Is this correct ?

My proof:

Firstly, C expands to fill the X-Y plane.....

n[(-1, 0), (+1, 0)] expands to a closed line segment to include the point (x, 0) for n > |x|.

The top unit disk is D = {v ∊ X-Y: ||v – (0, 1)|| ≤ 1}

2D is then = {2v ∊ X-Y: ||v – (0, 1)|| ≤ 1} = {2v ∊ X-Y: 2||v – (0, 1)|| ≤ 2}

2D = {2v ∊ X-Y: ||2v – (0, 2)|| ≤ 2} Rename the vector 2v as u (noting the bijection between 2v and v in X-Y, i.e. v ∊ X-Y ⇔ 2v ∊ X-Y), and then

2D = {u ∊ X-Y: ||u – (0, 2)|| ≤ 2}, i.e. the closed disk centred at (0, 2) with radius 2.

For a point in the upper open half-plane, (x, y) with y > 0, consider the expression $x^2 + y^2 – 2ny$.

With x and y fixed and y > 0, there is some n such that $x^2 + y^2 – 2ny < 0$

So, $x^2 + y^2 – 2ny + n^2 < n^2$

$x^2 + y^2 – 2ny + n^2 = x^2 + (y – n)^2 = ||(x, y – n)||^2 = ||(x, y) – (0, n)||^2 < n^2$

So, $(x, y) \in n.D$.

Similarly for the lower open half plane

Secondly for any ball $B_\epsilon(0) $ and any n, pick y such that $0 < y < \epsilon^2/4n$ and x such that $\epsilon^2/2 < x^2 + y^2 < \epsilon^2$ so that $(x, y) \in B_\epsilon(0)$

Then $\epsilon^2/2 – 2ny < x^2 + y^2 – 2ny $

$0 \le ε^2/2 – ε^2/2 < ε^2/2 – 2ny < x^2 + y^2 – 2ny$

So, $x^2 + y^2 – 2ny > 0$ and by the same steps as above, $||(x, y) – (0, n)||^2 > n^2$ and (x, y) is outside n.D.

Therefore n.D does not contain $B_\epsilon(0) $.