Here is an good example of a connected space which is not locally connected but I am stuck in disproving the local disconnectedness:
Let
$A$ $=$ $\big\{ (x, y)| x \in \mathbb{Q}^c $ and $0 ≤ y ≤ 1$ $\big\}$,and
$B$ $=$ $\big\{(x, y)|x \in \mathbb{Q}$ and $−1 ≤ y ≤ 0$ $\big\}$.
Clearly $A \cup B$ is a connected subset of $\mathbb{R}^2$ as the $x$-axis intersects with every upper and lower vertical line segment.
So the space will be the union of connected subspaces i.e. $\bigcup _{\alpha \in \mathbb{R} } {T}_\alpha$,
where for $\alpha \in \mathbb{Q}^c$ , ${T}_\alpha = \big\{ (x, y) | y=0 \big\} \cup \big\{ (\alpha, y)| 0 ≤ y ≤ 1 \big\}$, and
for $\alpha \in \mathbb{Q}$, ${T}_\alpha = \big\{ (x, y) | y=0 \big\} \cup \big\{ (\alpha, y)| -1 ≤ y ≤ 0 \big\}$
But how to prove that $A \cup B$ is not a locally connected subset of the Euclidean Space $\mathbb{R}^2$.
I was trying to show that any open set of this space can not be connected.
The precise statement we wish to prove is that there exists a point $p$ in our space $S := A \cup B$ and an open neighbourhood $U$ of $p$ such that for all open neighbourhoods $V$ of $p$ such that $V \subset U$, $V$ is disconnected.
Let's take $p = (0, -\frac 1 2 )$ and $U = \{q \in S : \| q - p\| < \frac 1 4 \}$. Notice that $U$ does not intersect the $x$-axis. So $U$ looks like a union of vertical line segments, which are not joined up.
Now all that remains is to show that any open neighbourhood $V$ of $p$ such that $V \subset U$ is disconnected. This is easy: just pick any irrational number $\alpha$ from the interval $(-\frac 1 4, \frac 1 4)$, and observe that $V$ is the union of the non-empty open subsets $\{ (x, y) \in V : x < \alpha \}$ and $\{ (x, y) \in V : x > \alpha \}$.