I'm looking for an example of a metric space $X$ such that for every $\epsilon > 0$ there exist connected subsets $A_1, \dots A_n$ for some $n \in \mathbb{N}$ such that $X = \cup_{i = 1}^nA_i$ and the diameter of each $A_i$ is strictly less than $\epsilon$ (this is called S-property) and that $X$ isn't locally connected (note that the sets $A_i$ aren't defined to be open or closed).
So far all I have been able to prove is that $X$ must not be compact because in other case these properties are equivalent but I haven't been able to come up with an example so any help would be appreciated.
As an edit I would like to add that I'm looking for an example of this because originally I was looking for an example of a metric space that has the S-property but isn't uniformly locally connected, where this means that for every $\epsilon > 0$ there exists a $\delta > 0$ such that if $d(x,y) < \delta$ ($d$ is the distance defined on $X$) then there is a connected set $C$ such that $x,y \in C$ and the diameter of $C$ is less than $\epsilon$. With this definition is clear that a space that is uniformly locally connected is locally connected and so I looked for a counterexample that wasn't locally connected but I'm also not sure if this is the correct way to go.
There can be no such space, i.e., the S-property implies local connectivity even without compactness.
To see this, suppose the S-property holds, let $x\in X$, and let $\epsilon>0$. Let $X = \cup_{i = 1}^n A_i$ be the given decomposition for diameter $\epsilon$. Note that we may assume WLOG that each $A_i$ is closed, since the closure of a connected set is still connected and has the same diameter.
Then we claim the component of $x$ in $B_\epsilon(x)$ contains some ball $B_\delta(x)$. Indeed, otherwise, there is a sequence $x_j\to x$ with each $x_j\in A_{i_j}$ for some $i_j$, and such that $x_j$ is not in the component of $x$ in $B_\epsilon(x)$.
Since there are only finitely many $A_i$'s, there is some index $i$ such that $i_j=i$ for infinitely many $j$'s, i.e., there is a subsequence $x_{j_k}$ completely contained in $A_i$ for some $A_i$. But then since $A_i$ is closed, we have $x\in A_i$, and since $\text{diam}(A_i)<\epsilon$, all of the $x_{j_k}$'s lie in the component of $x$ in $B_\epsilon(x)$, a contradiction.
Remark
For a counterexample to uniform local connectivity, note that the union of two open intervals $(-1,0)\cup (0,1)$ certainly has the S-property, yet is not uniformly locally connected, since $x$ and $y$ may be arbitrarily close and remain in different components.