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
Best Answer
Lemma: Let $X$ be a locally connected. If $U$ is an open set in $X$ then all connected components of $U$ are open sets in $X$.
Proof of the lemma: Let $U$ be an open set in $X$ and $W$ one component of $U$. For any $x\in W$, there is an open connected neighbourhood $W_x$ in $X$, such that $W_x\subseteq U$. But since $W_x$ is connected, we have $W_x\subseteq W$. So we have proved that, for any $x\in W$, there is an open neighbourhood $W_x$ such $W_x\subseteq W$. So $W$ is open.
Now let us answer the question:
Take $y \in Y$. Let $A$ be an open subset of $Y$ such that $y\in A$. Let $C$ be the component of $A$ such that $y\in C$. In order to prove that $C$ in an open connected neighbourhood of $y$ in $Y$, it is enough to prove $C$ is open.
Proof that $C$ is open:
Since the topology in Y is the quotient topology, we have that $f^{-1}(A)$ is open and that $f^{-1}(C)$ is a union of components of $f^{-1}(A)$. Since $X$ is locally connected, the components of open sets in X are open. So $f^{-1}(C)$ is a union of open sets in X. So $f^{-1}(C)$ is open. So $C$ is open in Y