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.)
Best Answer
It is compact and connected.
Your justificative for compactness is ok. A rigorous person could only ask why the set is closed and limited. For connectedness, you can argue that the set is path-connected. Given two points in there, you can find a path connecting them (hint: spherical coordinates, and think geometrically).