Connectedness can be a bit of an abstruse concept to work with. It's often easier to work with the stronger concept of path-connectedness (a space is path-connected if any two points can be joined by a continuous path in the space). Not every connected space is path-connected, but for those that are, this is generally the easiest way to prove connectedness.
In this case, for example, it's almost trivial to see that $X$ is path-connected: two points in the same disc can be joined by a straight-line path, while two points in different discs can be joined by a composite path formed of two line segments meeting at the tangency point.
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 $Y$.
Best Answer
We have to understand that connectedness must be defined in a metric (or topological) space. So, when we say "open" sets, we mean that they are open subject to a defined (metric) topology.
Simply, the connected (space) set $A \subseteq X$ is a set that is contained in a metric (or topological) space and there are not any two disjoint open (which implies the closed too) sets (in $X$) that make a partition for $A$. Namely, for any $U$ and $V$ non-empty open in $X$ and $U \cap V =\phi$, then $U \cup V \neq A$.
In conclusion, a set $A$ is connected if we can not find a partition for A from the open sets (subject to the topology (metric)).