Let $(X,d)$ be a metric space. Does the following property have a name?
For every $\epsilon>0$ a countable collection of subsets $(A_i)_{i=1}^\infty$ exists such that $X=\bigcup_{i=1}^\infty A_i$ and $diam(A_i)<\epsilon$.
EDIT 2: Is separability equivalent?
On
I believe a space satisfying your property is separable: Consider a for some fixed $\delta>0$, any point $x$ in your space. Now let $\epsilon<\delta$ and consider the centers of the given countable cover by $A_i$'s of your space. Since $x$ is in your space, the $A_i$'s cover the space, and $\epsilon<\delta$, it follows that there is an $i_0\in\mathbb{N}$ such that $x\in A_{i_0}$, moreover, we have that the center of $A_{i_0}$, $a_{i_0}$ must be within the $\delta$-neighborhood of $x$.
Note: There is one tricky point I haven't dealt with, and that's that nothing in the property you've given requires that the centers of the sequence of centers of the $A_i$'s for a smaller $\epsilon$ will contain the centers of a sequence for a larger $\epsilon$, but this can be addressed by 'appending' them to the sequence (we take add in the $A_i^\epsilon$'s to the sequence for countably many epsilons (e.g. use rational epsilons), and hence obtain a countable dense subset; it's still countable because countable union of countable sets is countable).
The property you stated is equivalent to separability of $X$ (which is equivalent to $X$ being second countable, since it is metric). Let's denote $B_\epsilon(x)$ the open ball of radius $\epsilon$ around $x$.
If $X$ is separable, take a countable dense subset $D=\left\{x_1,x_2,\ldots\right\}$. Given $\epsilon>0$, take $A_n=B_{\epsilon/2}(x_n)$. The collection $\left\{A_n\right\}$ has the desired properties.
Conversely, suppose the condition you wrote is satisfies. For a given $n$, choose a family $\left\{A^n_i\right\}_i$, with $diam A^n_i<1/n$ and $X=\bigcup_i A^n_i$. For each $n$ and each $i$ choose $x_{n,i}\in A^n_i$. Then $D=\left\{x_{n,i}:n,i=1,2,\ldots\right\}$ is a countable dense subset of $X$.