The problem is not with your understanding of divided, but rather with your understanding of closed. In the space $X=(0,1)\cup(2,3)$, the sets $(0,1)$ and $(2,3)$ are closed. This is because the topology $\tau$ on $X$ is the subspace (or relative) topology inherited from $\Bbb R$. A subset $U$ of $X$ is open in $X$ if and only if there is a $V\subseteq\Bbb R$ such that $V$ is open in $\Bbb R$ and $V\cap X=U$. Of course $(0,1)$ is open in $\Bbb R$, and $(0,1)\cap X=(0,1)$, so $(0,1)$ is open in $X$. By the definition of closed set this means that $X\setminus(0,1)$ is closed in $X$. And $X\setminus(0,1)=(2,3)$, so $(2,3)$ is closed in $X$. A similar argument shows that $(0,1)$ is also closed in $X$. Indeed, both of these sets are clopen (closed and open) as subsets of $X$, even though they are only open as subsets of $\Bbb R$. Openness and closedness depend not just on the set, but on the space in which it is considered.
You have the same problem with your first example: the sets $[0,1]$ and $[2,3]$ are clopen in the subspace $Y=[0,1]\cup[2,3]$ of $\Bbb R$, not just closed. For example, $[0,1]=\left(-\frac12,\frac32\right)\cap Y$, and $\left(-\frac12,\frac32\right)$ is open in $\Bbb R$, so $[0,1]$ is open in $Y$.
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
An example of such a space is $\Bbb Q$ with the relative topology. (I think) A general way to get such spaces easily is to consider a connected space and then consider a nice choice of dense subset and look at the relative topology. You could replace the rationals with the irrationals and it would also work.
I think another - more exotic - example is an irrational rotation set. Suppose $\theta\in(0,2\pi)$ is irrational, then let $S = \{e^{in\theta}:n\in\Bbb Z\}$. $S$ is dense in the unit circle but is not locally connected.