Let $(X,d)$ be a complete metric space and $n\in\mathbb N$. Suppose that every finite subset $F\subset X$ can be covered by $n$ closed balls of $X$ (that is, $N(Y,d,1)\le n$, in terms of covering numbers). Can $X$ itself be covered by $n$ closed balls? If not, what if $X$ is also assumed to be separable?
Some remarks
If $X$ is compact, it is true: $X$ is separable; if $S:=\{f_i\}_{i\ge1 }$ is dense in $X$ and for every $k$ $\displaystyle\{f_i\}_{1\le i\le k} \subset \bigcup_{1\le i\le n} \overline B(x_j^k,1)$, up to subsequences $(x_1^k,x_2^k,\dots,x_n^k)\to (x_1^\infty,x_2^\infty,\dots,x_n^\infty)\in X^n$ as $k\to\infty$, so $\displaystyle S\subset \bigcup_{1\le i\le n} \overline B(x_j^\infty,1) $ and then also $\displaystyle\overline S= X\subset \bigcup_{1\le i\le n} \overline B(x_j^\infty,1) $
If $X$ is not assumed to be complete, then it is not true in general: Here is an example: let $(e_i)_{i\ge1}$ be the standard Hilbert basis of $\ell_2$; let $u_i:=\frac1i\sum_{j=1}^ie_j$, let $X:=\{e_i\}_{i\ge1}\cup \{u_i\}_{i\ge1}\cup \{3ie_1\}_{1\le i< n}$, with the induced $\ell_2$ distance. Then we need $n-1$ unit balls to cover $\{3ie_1\}_{1\le i< n}$, and these balls are necessarily disjoint from the remaining points $\{e_i\}_{i\ge1}\cup \{u_i\}_{i\ge1}$, which however can’t be covered by a single unit closed ball centered at a point of $X$, because the only unit closed ball of $\ell_2$ containing every $e_i$ and every $u_i$ is the one centered at $0$, which is not in $X$.
If one uses open unit balls instead, in general it is false even for countable complete metric spaces: take the closure in $\ell_2$ of the preceding example, which is just $X\cup\{0\}$ (because $0=\lim_{i\to\infty}u_i$ is the only accumulation point of $X$). Then $X$ can’t be covered by $n$ open balls, fro the same argument.
