Let $(X,\tau)$ be the subspace of $\mathbb R^2$ consisting of the points in the line segments joining $(0,1)$ to $(0,0)$ and to all the points $(1/n,0)$, $n=1,2,3,\ldots$. Show that $(X,\tau)$ is connected but not locally connected.
A space is locally connected if every point admits a neighbourhood basis consisting entirely of open, connected sets.
I visualized this set, but I have no clue on how to prove both properties.
On
It's connected because it's path connected (any two points can be joined by a path passing through $(0,1)$). It's not locally connected at any point $(0,t)$ for $0\leq t<1$ because any such point has a basis of neighborhoods consisting of disjoint line segments limiting onto a vertical line segment. But it is locally connected everywhere else.
It is a connected space because each line segment joining $(0,1)$ to $(1/n,0)$ and the line segment joining $(0,1)$ to $(0,0)$ is connected and they share the point $(0,1)$.
It is not locally connected: consider the neighborhood $B(O,r)$ (where $0<r<1/2$) of $(0,0)$. You can find two separated line segments (from your space) in $B(O,r)$.