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 $Y$.
Take $U=(0,1)$ and $V=(2,3)$: these sets are both open and closed in the space $C$. $(0,1)$ are open in $C$ because each is the intersection with $C$ of a set open in $\Bbb R$, and each is closed in $C$ because it’s the complement in $C$ of an open subset of $C$. The fact that neither is closed in $\Bbb R$ is irrelevant.
Best Answer
The subspace topology on $Y \subseteq X$ is defined so that a set $S \subseteq Y$ is open iff there is an open set $U \subseteq X$ such that $S = U \cap Y$.
A set $Y \subseteq X$ is disconnected iff it is the union of two disjoint sets which are open in the subspace topology on $Y$.
You're missing the "which are open in the subspace topology" in your reasoning. In the subspace topology on $[0,1] \cup [3,4]$, the sets $[0,1]$ and $[3,4]$ are open because they are $(-1, 2) \cap ([0,1] \cup [3,4])$ and $(2, 5) \cap ([0,1] \cup [3,4])$, for instance.