Set of Points where X Fails to be Locally Connected I am stuck on a problem!  Suppose $X$ is a compact, connected metric space.  Let $L(X)$ be the set of points at which $X$ is not locally connected (here, locally connected means the point has a local basis of connected, open neighborhoods).  We call the weaker notion that $x$ has a neighborhood basis of connected - but not necessarily open - sets connected im kleinen, or cik.
Proposition: $L(X)$ has no isolated points.
I can't prove it.  It's false for compact metric spaces: Just take a sequence of points converging to $0$.  So the connectedness is necessary.  It is known that the set of points where $X$ fails to be cik has no isolated points.  
Anyone have an idea, or a reference?
 A: Got it.  If anyone sees a problem in the proof, let me know!  Let $N(X)$ be the set of points where $X$ is not cik.
Since the set of points where $X$ is cik has no isolated points and $N(X) \subset L(X)$, if $x \in L(X)$ is an isolated point then $X$ must be cik at $x$.  Let $V_n$ be a neighborhood basis (not necessarily open) of connected sets for $x$.  Since $x$ is isolated, it has a neighborhood $U$ such that $X$ is locally connected at every other point in $U$; pick an index $n$ so that $V_n \subset \overline{V_n} \subset U$.  For fixed $V_n$ each point of $\partial(V_n)$ has an open connected neighborhood contained in $W_\alpha \subset U$.
Then the union $Z_n$ of $V_n$ and these neighborhoods is connected, and it's open since $V_n = \text{int}(V_n) \cup \partial{V_n}$, i.e. it can be written as the union of the open $W_\alpha$ and $\text{int}(V_n)$.  Thus $Z_n \subset U$ is an open, connected neighborhood of $x$.  But $U$ was an arbitrary neighborhood of $x$, proving the proposition.
A: I think the first sentence of your proof is flawed. The rest is likely flawed as well, since you don't really use the assumption that $X$ is a continuum.
Suppose $p$ is an isolated point of $L(X)$. 
Then there is an open set $U$ such that $p\in U$, $X$ is locally connected at each point in $U\setminus \{p\}$, and $p$ has no connected open neighborhood in $U$.  
Since the component of $p$ in $U$ is not open, $p$ has no connected neighborhood in $U$ (so $X$ is not cik at $p$).
Now you want to just follow the proof of Theorem 3.4 in this paper.  The author uses weakly locally connected instead of cik.  The theorem shows that there is a continuum $C$ such that $p\in C\subseteq L(X)$ and $|C|>1$. Clearly $p$ is not an isolated point of $C$, so $p$ is not an isolated point of $L(X)$.
The basic idea is that in $\overline U$ there must be a sequence components $C_n$ such that $p\in \overline{\bigcup_{n<\omega} C_n}$ and $C_n\cap \partial U\neq\varnothing$ for each $n<\omega$. The limit of this sequence of continua is a continuum $C$ that contains $p$ and intersects $\partial U$. If $x\in C\cap U\setminus \{p\}$, then $x$ has no connected neighborhood in $U$ (the neighborhood would meet some $C_n$ and contradict its maximality).
