It is not locally connected at $p$ : for any bounded open neighborhood $U$ of $p$ its intersection $V$ with $0x$ is open and bounded, and let $v$ be its $\sup V$ and $n$ be the minimum index such that $a_n<v$ . Then $U$ may contain some points $a_i$ for $i>=n$ but it does not contain $a_{n-1}$ . At the same time it does contain some points of $((a_n,0),(a_{n-1},0))$ and, therefore, some points of the line segments $[(a_{n-1},0),(a_n,1/k)]$ for sufficiently large $k$ . Those parts of the set are disconnected and the neighborhood can be separated. (A more formal prove would require a more formal statement of the problem and some technical work.) Now, for any neighborhood of $p$ we may find an open ball $B$ such that its intersection with the set is contained in the neighborhood, then find some $m$ such that $a_m,a_{m+1}, . . .$ and their line segments are in the ball — this will be the connected subspace, which does contain another neighborhood of $p$ . The reason the previous exercise argument does not work here is that the set is not (weakly) locally connected at some points in any neighborhood of $p$ , and the components can’t be represented as the union of open sets.