$\{ \frac{1}{n} \mid n \in \mathbb{Z_+} \}$ is locally connected in $\mathbb{R}$ endowded with the usual topology, but its closure not Let $X = \{ \frac{1}{n} \mid n \in \mathbb{Z_+} \}$ and let us consider
$\mathbb{R}$ with its usual topology. I've proved that $X$ is locally connected by these means:

Let $x \in X$ and $U = (a,b)$ an open neighborhood of $x$. In this
topology, $U$ is also connected, since it's an interval. Thus, $U$ is
a connected neighborhood of $x$ contained in $U$.

However, I'm not sure if I should consider the subspace topology instead. Besides, to prove that $\overline{X}$ is not locally connected I think that $0$ must be the problem, but I don't know how to start.
 A: Obviously $\overline{X} = \{0\} \cup \{1/n : n \in \Bbb Z_+\}$. To show $\overline{X}$ is not locally connected, we need to "contradict" the definition of locally connected for $\overline{X}$.
Well, certainly $\overline{X}$ is open in itself (subspace topology must be considered) and obviously $\overline{X}$ contains $0$. We aim to show that every open set $U$, containing $0$ and residing in $\overline{X}$, fails to be connected. So let $U = (u, v) \cap \overline{X}$ where $u < 0 < v$. Obviously there is $N \in \Bbb Z_+$ (just use Archimedean property) s.t.
$$
1/(N + 1) < 1/N < v
$$ Now, we can use this to create a disconnection of $U$.
Indeed, take any real falling between $1/(N + 1) < \xi_N < 1/N$ (notice, by choice $\xi_N$ can never be of the form $1/n, n \in \Bbb Z_+$) Then
$$
U_1 = (u, \xi_N) \cap \overline{X} \text{ and } U_2 = (\xi_N, v) \cap \overline{X}
$$ are nonempty open sets of $\overline{X}$ that disconnect $U$. You can check these last assertions easily.
Thus, in summary there is some open set $V$ containing $0$ ($\overline{X}$ in this case), within which no open set $U$ with $0 \in U \subseteq V$ can be connected. So $\overline{X}$ cannot be locally connected at $0$.
