Local connection of closure of open connected subspaces in some metric spaces. Let $X$ be a metric, connected, locally connected and locally compact space, and let $U$ be an open connected subspace of $X$. Is $\overline{U}$ always locally connected?
I´m writing a thesis about Peano spaces and this would be useful if it were true, but I haven´t been able to prove it or find any counterexamples.
 A: Expanding on the previous answer, it's not true even if $U$ is connected and bounded and $X$ is a continuum.  Just let $X$ be a large, closed disc in the plane and let $U$ be a tubular neighborhood of sin$(\frac{1}{x})$ near $0$.
The spaces you want are the hlc spaces.  Look at examples of Peano curves that aren't hlc for more sophisticated obstructions.
A: No. Let $X=\mathbb{R}^2$ and define $U$ as follows:
$$U_n=\bigg(\frac{1}{n}-\epsilon_n,\frac{1}{n}+\epsilon_n\bigg)\times\mathbb{R}$$
We choose each $\epsilon_n>0$ so that closures of those subsets are pairwise disjoint, something like $\frac{1}{2n(n+1)}$.
Then we put
$$V=\mathbb{R}\times(-1,1)$$
$$U=V\cup\bigcup_{n=1}^\infty U_n$$
The role of $V$ is only to ensure that $U$ is connected. $U$ is also open but for $W=\{0\}\times\mathbb{R}$ we get
$$\overline{U}=\overline{V}\cup\bigcup_{n=1}^\infty \overline{U_n}\cup W$$
which is not locally connected (points at $W$ outside of $\overline{V}$ don't have small enough connected neighbourhoods).
This example can easily be modified to exclude $X$ compact as well. Simply consider $X=[-x,x]^2$ square for $x>1$ and then our open subset is $U\cap X$.
