Is intersection of refining locally connected topologies locally connected? Let $(X, \mathcal{T})$ be a topological space, $S(X)$ be the set of topologies in $X$, and
$$\mathcal{T}^P = \bigcap \{\mathcal{T}^* \in S(X) : \mathcal{T} \subset \mathcal{T}^* \textrm{ and } (X, \mathcal{T}^*) \textrm{ is locally connected}\}.$$
Is $\mathcal{T}^P$ locally connected?
Notes:

*

*Discrete topology refines any topology and is locally connected, so the above intersection is well-defined.

*$\mathcal{T}^P$ is a topology in $X$ (proof here)

*$\mathcal{T} \subset \mathcal{T}^P$

*If $(X, \mathcal{T})$ is locally connected, then $\mathcal{T}^P = \mathcal{T}$.

*If $(X, \mathcal{T})$ is finite or finitely generated, then $(X, \mathcal{T})$ is locally connected.

*So a possible counter-example must have infinite $X$.

*Quotient map preserves local connectedness.

It seems to me that there should be a counter-example.
 A: Yes.  Given a topology $\mathcal{T}$ on $X$, let $c(\mathcal{T})$ be topology in $X$ generated by the $\mathcal{T}$-connected components of all the $\mathcal{T}$-open subsets of $X$.  Then any locally connected topology $\mathcal{T}^*$ containing $\mathcal{T}$ must contain $c(\mathcal{T})$, since if $U$ is $\mathcal{T}$-open then each $\mathcal{T}$-connected component of $U$ is a union of $\mathcal{T}^*$-connected components of $U$ which are $\mathcal{T}^*$-open by local connectedness.  Also, if $\mathcal{T}=c(\mathcal{T})$, then $\mathcal{T}$ is locally connected, since the connected components of each element of $\mathcal{T}$ are in $c(\mathcal{T})$.
Now iterate this construction transfinitely.  That is, recursively define $c^\alpha(\mathcal{T})$ for each ordinal $\alpha$ by $c_0(\mathcal{T})=\mathcal{T}$, $c^{\alpha+1}(\mathcal{T})=c(c^{\alpha}(\mathcal{T}))$, and $c^\alpha(\mathcal{T})$ is the topology generated by $\bigcup_{\beta<\alpha}c^\beta(\mathcal{T})$ if $\alpha$ is a limit ordinal.  This defines an increasing sequence of topologies on $X$ which must eventually stabilize (since there are more ordinals than topologies on $X$).  That is, for some $\alpha$, $c(c^\alpha(\mathcal{T}))=c^\alpha(\mathcal{T})$.  Then $c^\alpha(\mathcal{T})$ is locally connected.  But also, any locally connected topology containing $\mathcal{T}$ contains $c^\beta(\mathcal{T})$ for all $\beta$ (by induction on $\beta$).  It follows that your $\mathcal{T}^P$ is equal to $c^\alpha(\mathcal{T})$ and is locally connected.
