Cardinality of a minimal open cover of the disc 
Consider $D_1^2(0)=\{x\in\Bbb R^n: ||x||_2\leq 1\}$ and let $\epsilon>0$. Consider the open cover $\mathcal{O}=\{B_\epsilon^2(x):x\in D_1^2(0)\}$ of $D_1^2(0)$. What is the minimum cardinality of a minimal subcover of $\mathcal{O}$?
If this seems intractable, can the upper bound $(\lceil \sqrt{n}/\epsilon\rceil+1)^n$ be improved?

I am able to calculate an upper bound of $(\lceil\sqrt{n}/\epsilon\rceil+1)^n$ by considering the 'obvious' open covering of $D_1^\infty(0)=\{x\in\Bbb R^n: ||x||_\infty\leq 1\}$ where each element has radius $\epsilon/\sqrt{n}$ and then considering the image of this open covering and $D_1^\infty(0)$ itself under the map $$f(x)=\begin{cases}\frac{||x||_2}{||x||_\infty} x&\text{ if } x\neq0\\0&\text{ if }x=0\end{cases}$$
which is a Lipschitz map with Lipschitz constant $\sqrt{n}$ from $(\Bbb R^n, ||\cdot||_\infty)$ to $(\Bbb R^n, ||\cdot||_2)$. I'm sure some easy improvements could be made by more careful considerations of $D_1^\infty(0)$, but improvements in the constant $\sqrt{n}$ would be more helpful.
An easy lower bound for this cardinality is $\lceil 1/\epsilon^n\rceil$ by measuring arguments, and this approximation is attained for $\epsilon$ near $1/2$. But I suspect that this lower bound becomes strict for small $\epsilon$ since the Lebesgue covering dimension of $B_1^2(0)\subseteq \Bbb R^n$ is $n$, and overlaps are going to become more frequent the smaller $\epsilon$ becomes.
Any help is appreciated.
 A: Recall that the $n$-dimensional volume of a ball of radius $r$ is $\pi_nr^n$ with
$$\tag1 \pi_n=\frac{\pi^{n/2}}{\Gamma(\frac n2+1)}=\begin{cases}\frac{\pi^k}{k!}&\text{if $n=2k$}\\
\frac{k!2^n\pi^k}{n!}&\text{if $n=2k+1$}\end{cases}$$
(where $\pi_1=2$, $\pi_2=\pi$, $\pi_3=\frac43\pi$, etc.). Note that $\pi_n\to 0$ as $n\to \infty$. More specifically, using Stirling's formula we obtain
$$\tag2\sqrt[n]{\pi_n}\approx {\sqrt {\frac{2\pi e}{n}}}\cdot \frac1{\sqrt[2n]{\pi n}}\approx \sqrt {\frac{2\pi e}{n}}.$$
(A closer examination shows that $\approx$ can be replaced with $<$, which is in favour of our estimations done below)
For $r>0$, Let $f_n(r)$ denote the minimal number of open balls of radius $1$ with center in the open ball of radius $r$ needed to cover the latter. This corresponds to the number you are looking for in your question with $\epsilon:=\frac1r$.
Note that the restriction that all centers of the $1$-balls are inside the $r$-ball can be dropped (provided $r>1$): If a $1$-ball with does not even intersect the $r$-ball it can simply be dropped; and if its center is not interior but it intersects the $r$-ball, then their boundaries intersect in a hyperplane; if we reflect the $1$-ball at this hyperplane, it has center in the interior of the $r$-ball and covers strictly more of it.
From comparing the volume, we obtain the seemingly very weak lower bound
$$\tag3f_n(r)\ge r^n.$$
Consider an unimodular lattice $\Lambda\subset\mathbb R^n$ (that is, the volume of a fundamental domain is $1$). Let $U$ be a fundamental domain of $\Lambda$ and assume $U\subseteq \overline{B^2_\rho(u_0)}$ for some point $u_0$. Then any point in $\mathbb R^n$ is at distance $\le\rho$ from a lattice point. Hence for $a<\frac1\rho$, the $1$-balls with centers in $a\Lambda$ cover the whole space. To cover the $r$-ball it suffices to take those centers contained in $a\Lambda\cap B_{r+1}(0)$. For each $x\in a\Lambda\cap B_{r+1}(0)$ consider $aU+x$. These are pairwise disjoint, are contained in $B_{r+1+a\rho}(au_0)$ and have volume $a^n$ each. Hence the number of such latticepoints can be estimated from above and gives us an upper bound
$$f_n(r)\le \frac{\pi_n(r+1+a\rho)^n}{a^n}  $$
and in the limit as $a\to \frac1{\rho}^-$, 
$$ f_n(r)\le \pi_n\cdot\left(\rho (r+1)+1\right)^n=\left(\sqrt[n]{\pi_n}\rho(r+1)+\sqrt[n]{\pi_n}\right)^n.$$
The number $\sqrt[n]{\pi_n}\rho$ corresponds to the $\sqrt n$ in your formula.
Remains to make a good pick for $\Lambda$ with small $\rho$.
The simplest choice is $\Lambda=\mathbb Z^n$, $U=[0,1)^n$, $u_0=(\tfrac12,\ldots,\tfrac12)$, $\rho =\frac12\sqrt n$. Using the approximation $(2)$, this gives us an improvement from your $\sqrt n$ to 
$$\tag4\sqrt[n]{\pi_n}\rho\approx \sqrt {\frac{2\pi e}{n}}\cdot \frac12\sqrt n=\sqrt{\frac{\pi e}{2}}\approx 2.066.$$
(According to the remark above, the $\approx$ are can in fact be replaced with $<$)
Combining $(3)$ and $(4)$ we have
$$ r\le{\sqrt[n]{f_n(r)}}\le \sqrt{\frac{\pi e}{2}}\cdot r+O(n^{-1/2})$$
The morale of this is that in higher dimensions, balls are not as fat as they look in up to three dimensions.
We can try to improve $(4)$ by using the hexagonal lattice in dimension $2$, or the $E_8$ lattice in dimension $8$, or the Leech lattice in dimension $24$ as building blocks. This is left as exercise for the reader.
