Locally connected Locally compact separable metric space

Question

Let $X$ be a locally connected locally compact separable metric space. Is it possible to find a countable collection $\mathcal{B}$ such that every member of $\mathcal{B}$ is a nonempty peano subspace of $X$ and $\cup \mathcal{B}=X$.

Answer 1

For each point $p \in X$, let $F(p)$ be the set of neighborhoods of $p$, where, if $f$ is in $F(p)$, then $f$ is connected with compact closure, and $f$ is a subset of some $g$, where $g$ also belongs to $F(p)$. Note that $F(p)$ is nonempty.

Let $G(p) = \bigcup F(p)$. We can show (I leave this to you) that $G(p)$ is connected and that if $G(p)$ intersects $G(q)$ for any $q \in X$, then $G(p)=G(q)$.

The collection of all such $G(p)$ is therefore an open partition of $X$, and since $X$ is separable, it is also ccc. (The set of all $G(p)$ is countable.) So if each $G(p)$ is a countable union of Peano spaces, then $X$ is also. But a separable metrizable space is hereditarily Lindelöf, and $F(p)$ is an open cover of the subspace $G(p)$, so $F(p)$ has a countable subset $H(p)$ such that $\bigcup H(p)= \bigcup F(p)=G(p)$. And each member of $H(p)$ is compact and connected.