Suppose $C$ is a component of $X$, where $X$ is locally connected. If $x \in C$, then $x$ has a connected neighbourhood $U_x$. So $C \cup U_x$ is connected, as the union of two intersecting (in $x$) connected sets and $C \subseteq C \cup U_x$. So by maximality $U_x \subseteq C$. So $x$ is an interior point of $C$, and as $x \in C$ was arbitrary, $C$ is open. A connected component is always closed (or $\overline{C}$ would be a strictly larger connected subset). So $C$ is clopen.
This means that if $C$ is a component of $x$ in $X$, $C$ is one of the clopen subsets we intersect in computing the pseudocomponent. So the pseudocomponent $P_x$ of $x$ is a subset of $C$. By general theory, which I hope you have covered, the component of $x$ is a subset of the pseudocomponent $P_x$ of $x$ and so we have equality.
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
The fact that $X$ is locally connected iff for all open subsets $O$ of $X$, all components of $O$ (as a subspace) are open in $O$ (and $X$ too), should be known to you. It's Munkres theorem 25.3
So we will show $X$ to be locally connected using the right to left direction of this theorem.: let $O$ be open in $X$ and let $C$ be a component of $O$ (in the subspace topology).
Let $x \in C$. Then $x \in O$ and $O$ is a neighbourhood of $x$ so by assumption of weak locally connectedness there is a connected set $C_x \subseteq O$ such that $x \in \operatorname{int}(C_x)$ (as $C_x$ is a a neighbourhood of $x$).
Then $C$ and $C_x$ are both connected subsets of $O$ that intersect (in $x$) so $C \cup C_x$ is also a connected subset of $O$. As $C$ is a component of $O$ and as such is a maximally connected subset of $O$,
$$C \cup C_x = C \text{ so } x \in \operatorname{int}(C_x) \subseteq C$$
showing finally that $x$ is an interior point of $C$. As this holds for all $x \in C$, $C$ is open, as required.