Show that $X$ is not locally connected at $p$

Your part on connectedness is quite ok, but it needs some clarification.

X is not locally connected

Because of the definition of $X$ as the space in the picture, it can be assumed that $X$ is in the real plane and has the subspace topology.

Consider a (small) open neighbourhood of $p$, so an open ball $B$, such that there exist $n \in \mathbb{N}$ with $a_n$ not in $B$, but $a_{n+1}$ is in $B$, then because the ball is open there will be some part of some stalks of the $n$th-broom lying in $B$, this implies the non-local connectedness of the space.

Remark: We want a small neighbourhood because we want to show a local property, then the fact that such $n$ exists is a tautological statement about what an open ball is

X is weakly connected in $p$

We have to show that $X$ is weakly connected in $p$

To be weakly connected in $p$ means that given an open neighbourhood of $p$, I can find a subset of this neighbourhood such that $p$ is in the interior of this subset, and this subset is connected.

So, the difference with local connectedness is that this subset needs not to be open.

Given a small neighbourhood $B_\epsilon$ of $p$ of radius $\epsilon$, then there exist $N \in \mathbb{N}$ such that $a_n \in B_\epsilon$ for all $n > N$, then I can take the subspace of $X$ including $p$ and every broom until the $n$th-one , $n>N$, this space is (obviously not open but) connected and its interior contains $p$, so $X$ is weakly connected in $p$.

Remark: Note that an interior point is with respect to $X$, so using the induced topology