Set of Points where $X$ is locally connected is a $G_\delta$-set I had this question on my final:

Let $(X,d)$ be a metric space. Prove that the set of points where $X$ is locally connected is the countable intersection of open subsets of $X$.

I wasn't quite sure how to approach the problem; I know that the components of open subsets of $X$ are open if $X$ itself is locally connected, but I wasn't sure where to go with the information (if anywhere at all). Any hints? If possible, I'd prefer hints to begin with so I can work out a solution. After I have solved it, anyone can return to edit their solution to be more complete so as not to detract from the value of MSE as a reference site.
 A: Note first that contrary to the claim in comments, this set (call it $LC(X)$) need not be open.
Let $X \subset \mathbb{R}^2$ be the union of all the lines through the origin that have rational slope, with the Euclidean metric.  $X$ is locally connected at the origin 0; indeed any open ball in $X$ centered at 0 is path connected.  However, for any other point $y \in X$, any neighborhood of $y$ which is so small that it does not contain 0 is not connected (it contains a point on some other line), so $X$ is not locally connected at any other point.  Thus $LC(X) = \{0\}$, which is certainly not open in $X$.
To show in general that $LC(X)$ is $G_\delta$, here's a hint: for any $R > 0$, let $LC_R$ be the set of all $x \in X$ such that there exists an $r_x < R$ and a connected open set $U_x$ with $x \in U_x \subset B(x,r_x)$.  Show that $LC_R$ is open and that $LC(X)$ is the intersection of countably many of the $LC_R$.  (There might be a simpler way but this is the first thing that occurred to me.)
A: Let $LC(X)$ denote the set of points where $X$ is locally connected. If $LC(X)=\varnothing$, it clearly is a $G_\delta$-set as $\varnothing=\bigcap\varnothing$ and $\varnothing$ is open. 
Suppose $LC(X)$ is nonempty. Then there is a point $x\in LC(X)$ such that for every open set containing $U$ such that $x\in U$, there is a connected open set $V\subseteq U$ such that $x\in V$. Note then, that every point in $V$ will be in $LC(X)$ since if $y\in V$, then any open set $U'$ with $y\in U'$ intersects $V$ and $V\cap U'\neq\varnothing$, thus it is an open set containing $y$, in which case we can take an open ball around $y$ small enough so that it is contained in the intersection and is connected. This says that we found an open set $V\subseteq LC(X)$ around an arbitrary point $x$, hence $LC(X)$ is open. Hence, $$LC(X)=\bigcap \{LC(X)\}.$$ Does the argument sound complete? It seems to me that it is, but I'm not going to claim without any kind of uncertainty that it is flawless.
