Firstly, notice that this doesn't hold for an annulus or donut shape. You take two pieces that go around different sides of the hole and you're done. So somehow the structure of $I^2$ must be relevant (apparently its genus).
So without using more properties of the space, we're going to be stuck. Try drawing some pictures.
Here's one intuitively based approach: let $W$ be $(U\cap V)^c$, the closed subset of $I^2$ not in the intersection. Then $W=U'\cup V'$ where primes indicate restriction to $W$, and $U'\cap V'$ is empty. Hence either one of the primed sets is empty (and we're done as the sets were nested) or $W$ is disconnected.
In this case, use the simply connected property of the original space to note that, picking two random connected components $A,B$ of $U\cap V$, and points $a,b$ within them, any path joining $a,b$ can be continuously deformed into any other. The restriction of these paths to $W$ can also be continuously deformed into each other. But since $W$ is a disconnected union of the primed sets, you can show that each connected component is entirely surrounded by one of the primed sets, contradicting the connectedness of the other.
Edit: Just had a chat with a smarter friend than me, and we concluded that the 'natural' way to do this is homology theory. The logic runs roughly as follows:
$$0 \longrightarrow H_0(U\cap V) \longrightarrow H_0(U) \oplus H_0(V) \longrightarrow H_0(U\cup V) \longrightarrow 0$$
is a short exact sequence; since $U,V,U\cup V$ are connected, all their $H_0$s are $\mathbb Z$. But then since the sequence is exact (and in particular we have an injection for the second arrow) and since the kernel of the addition map $\mathbb Z \oplus \mathbb Z \to \mathbb Z$ is isomorphic to $\mathbb Z$, we conclude that $H_0(U\cap V) \cong \mathbb Z$ and therefore $U\cap V$ is also connected.
Simplifying this argument by stripping out all the general homological algebra would probably result in basically approximating $U,V$ by unions of simplices (with a deeply tedious argument) and then doing some (Euler-characteristic-style) counting to show that the intersection must be connected.
Note: One thing that's worth pointing out is that both the above arguments move to path-connectedness which is in general stronger than connectedness. However open subsets of $\mathbb R^n$ have the property that they are path connected if and only if they're connected. (Proof: Consider path components; show they are both closed and open. This contradicts connectedness.)
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
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.