My Topology teacher told us that the existence of a connected neighborhood of a point $x$ in a Topological space does not imply local connectedness at that point, that is, it is not true in general that there is a neighborhood base of connected neighborhoods of $x$.
He told us that we could get an example of this fact by making some "adjustment" or applying a certain "transformation" on the subspace $E=\{(\frac{1}{n},0)\in \mathbb{R}^2: n \in \mathbb{N}\} \cup \{(0,0)\}$ of the plane $\mathbb{R}^2$ (with the Euclidean subspace topology). I know that $E$ is locally connected at every point of the form $(\frac{1}{n},0)$, since every singleton $\{(\frac{1}{n},0)\}$ is open and connected in $E$, so it generates a connected neighborhood base at $(\frac{1}{n},0)$ whose only element is $\{(\frac{1}{n},0)\}$. On the other hand, $E$ is not locally connected at $(0,0)$, because every neighborhood of this point is disconnected.
Now, I don't know what to do to generate a space to exemplify the said fact. I suppose I should find a way to get some connected neighborhood of $(0,0)$ such that the collection of the connected neighborhoods don't form a base, but I really don't have good ideas on how to proceed. Any thoughts? Thanks on advance.
HINT: $\big(E\times[0,1]\big)\cup\big([0,1]\times\{1\}\big)$ (draw a picture) is connected, so every point of it has a connected nbhd, but it is not locally connected at the origin.