Snake-like continuum vs Peano continuum A snake-like continuum is a continuum such that for every $\varepsilon >0$ there exist a collection of open sets $d_1,d_2,\ldots d_n$ with diameters les than $\varepsilon$ such that $d_i \cap d_j\neq \emptyset$ iff $|i-j|\leq 1$ and $X=\cup_{i=1}^n d_i$.
A Peano continuum is a continuum such that for every $\varepsilon >0$ there exist a collection of connected sets $d_1,d_2,\ldots d_n$ with diameters les than $\varepsilon$ such that  $X=\cup_{i=1}^n d_i$.
Triod is an example of a Peano continua that is not snake-like.
My question is: Are all snake-like continua a Peano continua too? If the answer is affirmative, it is enough to prove that any open set can be represented as a finite union of connected sets (is that true for metric spaces?). Otherwise we should find a counterexample.

Since you insist of having a continuum, here is a counter-example:
Let $X$ be the subset of $R^2$ which is the union of the graph of the function
$$
y=\sin(1/x), 0<x\le 1
$$
and the vertical interval $\{(0,y): -1\le y\le 1\}$.

I leave it to you to check that $X$ is a snake-like continuum. It is not a Peano continuum since it is not even path-connected.
However, I do not have examples of path-connected snake-like  continua which are not Peano. There is a good chance, they do not exist.
