I am having trouble proving a result from a paper, which of course includes no proof. I wonder if the author had a simple - but flawed - argument in mind, or if I'm just being a dunce. It is Theorem 3 here: https://www.jstor.org/stable/2372339
It involves the following property:
If $X$ is a metric space and $x, y \in X$, then we say that $X$ is aposyndetic at $x$ with respect to $y$ if there is a compact, connected neighborhood of $x$ not containing $y$. Let $A(x)$ be the set of points $y$ such that $X$ is not aposyndetic at $y$ with respect to $x$ - that is to say, the points $y$ all of whose closed, connected neighborhoods also contain $x$.
The following is what I'm trying to prove: If $X$ is a compact, connected metric space and $x \in X$ is a point, then $A(x)$ is connected.
In fact it's also closed, which I can show. Maybe it is something very simple, but I'm just not seeing it. It may be a useful fact that the nested intersection of compact, connected subsets of $X$ is also connected. Does anyone see the proof?