Locally connected iff every open cover has a connected refinement? Is it true that a space is locally connected iff every open cover has a refinement into connected open sets? I know locally connected implies this, but I'm not sure about the converse. I'm having trouble thinking of a counterexample.
 A: The converse is true at least for regular spaces. Suppose that $X$ is regular, and every open cover of $X$ has an open refinement by connected sets. Let $x\in X$, and let $U$ be any open nbhd of $x$. By regularity there is an open $V$ such that $x\in V\subseteq\operatorname{cl}V\subseteq U$. Let $\mathscr{W}$ be an open refinement of the open cover $\{U,X\setminus\operatorname{cl}V\}$ by connected sets. There must some $W\in\mathscr{W}$ such that $x\in W$, and $W\nsubseteq X\setminus\operatorname{cl}V$, so $W$ is a connected open nbhd of $x$ contained in $U$.
I’ll have to think about possible non-regular counterexamples. (Note that regular here does not imply Hausdorff.)
Added: It is not true in general. For $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k\ge n\}$, and let $Z$ denote $\Bbb N$ with the topology
$$\{\varnothing\}\cup\{U_n:n\in\Bbb N\}\;.$$
Let $K=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the topology that it inherits from $\Bbb R$. Let $Y=Z\times K$ with the product topology, and let $X=Y/(\{0\}\times K)$. Let $p$ be the point of $X$ corresponding to $\{0\}\times K$.
The only open nbhd of $p$ is $X$ itself, so $X$ is connected: it has no non-empty proper clopen subset. This also implies that every open cover of $X$ has an open refinement by connected sets, namely, $\{X\}$; indeed, it even has a subcover by connected sets.
However, the open nbhd $V=U_1\times K$ of $\langle 1,0\rangle$ does not contain a connected open nbhd of $\langle 1,0\rangle$: every open nbhd of $\langle 1,0\rangle$ contained in $V$ contains a proper open subset $W=U_1\times\left\{\frac1n\right\}$ for some $n\in\Bbb Z^+$, and $W$ is clopen in $V$, so $\langle 1,0\rangle$ does not have a local base of connected open nbhds, and $X$ is not locally connected.
$X$ is $T_0$ but not $T_1$ or regular.
A: Property $P$
Let $(X, \mathcal{T})$ be a space and $P$ be the following property:
for each $x \in X$ and non-connected $U_x \in \mathcal{T}(x)$ there exists $V_x \in \mathcal{T}$ such that $U_x \cup V_x = X$ and $x \not\in V_x$.
Here $\mathcal{T}(x)$ denotes the set of open neighborhoods of $x$.
Theorem
Suppose $X$ has property $P$ and each open cover has an open connected refinement. Then $X$ is locally connected.
Proof
Let $x \in X$ and $U_x \in \mathcal{T}(x)$. If $U_x$ is connected, then $U_x$ is trivially a connected open subset contained in $U_x$. Suppose $U_x$ is not connected. By assumption, there exists $V_x \in \mathcal{T}$ such that $U_x \cup V_x = X$ and $x \not\in V_x$. Also by assumption, there exists an open connected refinement of $\{U_x, V_x\}$. Since $x \not\in V_x$, that refinement contains an open connected $W_x \in \mathcal{T}(x)$ such that $W_x \subset U_x$. Therefore $X$ is locally connected.
Theorem
Let $(X, \mathcal{T})$ be an $R_0$ space. Then $X$ has property $P$.
Proof
Let $x \in X$ and $U_x \in \mathcal{T}(x)$. Let $[x] \subset X$ denote those points of $X$ which have the same neighborhoods as $x$. By assumption, $[x]$ is closed in $X$. Let $V_x = X \setminus [x]$. Then $V_x \in \mathcal{T}$, $x \not\in V_x$, and $U_x \cup V_x = X$. Therefore $X$ has property $P$.
Note
Each regular space is $R_0$ (here regular does not necessarily imply Hausdorff).
Counter-example for general converse
Here is a counter-example for a general converse. It uses the topology I recently learned from here. Let $X = \mathbb{N}$, and $\mathcal{T}$ consist of the following sets:

*

*$\emptyset$,

*$X$,

*those that contain $1$, are cofinite, and do not contain $0$,

*$\mathcal{P}(\mathbb{N}^{\geq 2})$, where $\mathcal{P}$ denotes subsets.

Then $\mathcal{T}$ is a topology on $X$. Any open cover of $X$ contains $X$, since it is the only way to cover $0$. $X$ is also connected for the same reason. Therefore every open cover of $X$ has an open connected refinement. However, $X$ is not locally connected at $1$, since each of its open neighborhoods is disconnected.
As a sanity check, $X$ is not $P$. For pick any $U_1 \in \mathcal{T}(1)$ and $V_1 \in \mathcal{T}$ such that $U_1 \cup V_1 = X$. Then $V_1 = X$, since that is the only way to cover $0$.
Strengthening for equivalence
The following strengthening is clearly equivalent to local connectedness: every open subspace has the property that every open cover has an open connected refinement.
