Let $X$ be a connected compact metric space. Given a positive $\varepsilon$ and two points $x,y\in X$ we write $x\sim\_\varepsilon y$ if there exists a sequence $C\_1,\dots,C\_n$ of connected subsets of diameter $<\varepsilon$ in $X$ such that $x\in C\_1$, $y\in C\_n$ and $C\_i\cap C\_{i+1}\ne\emptyset$ for all $i<n$. It is clear that $\sim\_\varepsilon$ is an equivalence relation on $X$. The equivalence class $[x]\_\varepsilon:=\{y\in X:y\sim\_\varepsilon x\}$ will be called the *$\varepsilon$-connected component* of $x$.


**Problem.** What can be said about the $\varepsilon$-connected components of $X$? In particular, is each $\varepsilon$-connected component $[x]\_\varepsilon$ dense in $X$? Is each $\varepsilon$-connected component $\sigma$-compact?


**Added in Edit.** It can be shown that each $\varepsilon$-connected component $[x]\_\varepsilon$ is $\sigma$-compact. To prove this fact, fix a countable base $\mathcal B$ of the topology of $X$, which is closed under finite unions and let $\mathcal B\_\varepsilon$ be the subfamily of $\mathcal B$ consisting of basic sets of diameger $<\varepsilon$. It can be shown that each connected subset of diameter $<\varepsilon$ is contained in the connected component of some set $\bar B$ with $B\in\mathcal B\_\varepsilon$. Now fix an enumeration $\{B\_n\}\_{n\in\omega}$ if the family $\mathcal B\_\varepsilon$. Given any point $x\in X$ let $C\_0=\{x\}$ and for every $n\in\mathbb N$ let $C\_n$ be the union of $C\_{n-1}$ with all connected components of $\bar B\_n$ that intersect $C\_{n-1}$. It can be shown that the sets $C\_n$ are compact and $[x]\_\varepsilon=\bigcup\_{n\in\omega}C\_n$.



Are $\varepsilon$-connected components dense?
I believe $[x]\_\varepsilon$ is always $\sigma$-compact. First consider a slightly modified definition where we replace 'diameter $<\varepsilon$' with 'diameter $\leq \varepsilon$' and call the corresponding set $[x]\_{\leq\varepsilon}$. 


Note that $[x]\_\varepsilon = \bigcup\_{\delta < \varepsilon}[x]\_{\leq\delta}$, since being in $[x]\_\varepsilon$ is witnessed by a finite sequence of sets whose diameters must have a uniform bound below $\varepsilon$. So we also have $[x]\_\varepsilon = \bigcup\_{n}[x]\_{\leq(\varepsilon -2^{-n})}$ so if we can show that $[x]\_{\leq\varepsilon}$ is always $\sigma$-compact, then we'll have that $[x]\_\varepsilon$ is always $\sigma$-compact as well.


So the second claim, as mentioned in the comments, is that we can restrict our attention to closed $C\_i$ in the definition. This is because the closure of a connected set is connected and has the same diameter.


Fix a $\varepsilon > 0$. For any closed set $F$, let $F^\ast = \bigcup\{G:G\text{ closed, connected, }G\cap F\neq \varnothing\text{, diam}(G)\leq\varepsilon\}$. Now notice that $\{G:G\text{ closed, connected, }G\cap F\neq \varnothing\text{, diam}(G)\leq\varepsilon\}$ is a closed subset of hyperspace (connectedness, having non-empty intersection with $F$, and having diameter $\leq \varepsilon$ are all closed conditions in hyperspace). This implies that $F^\ast$ is also a closed set, since the union of a closed family of sets is closed (since $X$ is compact). 


Then some definition chasing tells us that $[x]\_{\leq\varepsilon}$ is $\{x\}\cup\{x\}^\ast\cup\{x\}^{\ast\ast}\cup\dots$, so that in particular $[x]\_{\leq \varepsilon}$ is a $\sigma$-compact set.


Therefore $[x]\_\varepsilon$ is $\sigma$-compact as well.


As for the density question, I suspect that you can use the boundary bumping theorem to show that $[x]\_\varepsilon$ always contains the composant of $x$, which would imply that it's always dense since composants are always dense.


The topologist's sine curve gives an example of a non-dense $\varepsilon$-connected component. Specifically the $\varepsilon$-connected component of any point in the line segment is not dense for sufficiently small $\varepsilon$.


