Space $X$ is called locally connected if it has a basis consisting of connected sets.
It's called second-countable if it has a countable basis.
If $X$ is both locally connected and second-countable, does that imply that there exists a countable basis consisting of connected sets?
Note that this is true if $X$ is a metric space, since for any basis $\mathcal{B}$ of $X$, we can just take a refinement of $\mathcal{B}_n = \{B\in\mathcal{B} : \text{diam}(B) < \frac{1}{n}\}$ consisting of connected sets from local connectedness, use the Lindelof property, and then the union of those covers for each $n$ will be a countable basis consisting of connected sets.
Best Answer
More generally, if $B$ and $C$ are two bases for a topological space $X$, then there is a basis $C_0\subseteq C$ of cardinality at most $|B|^2$ (so in your case, let $B$ be a countable basis and let $C$ be the connected open sets). To construct $C_0$, for each $U,V\in B$, fix an element $W_{U,V}\in C$ such that $U\subseteq W_{U,V}\subseteq V$, if any such element exists. Then I claim the set $C_0$ of all these $W_{U,V}$s is a basis as well.
To prove this, suppose $x\in X$ and $V$ is a neighborhood of $x$, which we may assume is an element of $B$; we wish to find an element of $C$ which contains $x$ and is contained in $V$. Since $C$ is a basis, there exists $W\in C$ such that $x\in W\subseteq V$, and then there also exists $U\in B$ such that $x\in U\subseteq W$. But then $W_{U,V}$ exists and is an element of $C_0$ such that $x\in W_{U,V}\subseteq V$.